qu.1.topic=Ch 2, Sec 1 - Displacement, Position, and Distance@ qu.1.1.mode=Randomized Formula@ qu.1.1.editing=useHTML@ qu.1.1.name=A Person Walks@ qu.1.1.question=$name $does $north m north. After that he turns around and $does $south m south. Find $parameter@ qu.1.1.answer=$ans m (1 ? 0.025)@ qu.1.1.random= $integer\=rint(2); $name\=switch(rint(8), "John", "Mike", "Fred", "Joe", "Tom", "Derek", "Dan", "James"); $does\=switch(rint(4), "walks", "jogs", "runs", "drives"); $apostr\="'s"; $north\=rand(10,100,3); $south\=rand(10,100,3); $parameter\=switch($integer, "$name$apostr displacement. Assume the x axis to be directed north.", "the total distance $name covered."); $calculation\=switch($integer, $north-$south, $north+$south); $ans\=sig(3,$calculation)@ qu.1.1.hint.1=Displacement is characterized by magnitude and direction. Displacement in the direction of the coordinate axis is positive, and displacement in the direction opposite to the direction of the coordinate axis is negative.@ qu.1.1.comment= 1. Displacement is characterized by magnitude and direction. Displacement in the direction of the coordinate axis is positive, and displacement in the direction opposite to the direction of the coordinate axis is negative.

2. Displacement equals to the difference between the final and the initial coordinates of an object: Dx = x_final - x_initial.
Total distance traveled simply equals to the sum of distances traveled in each direction. @ qu.1.2.mode=Randomized Formula@ qu.1.2.editing=useHTML@ qu.1.2.name=Practicing on Track@ qu.1.2.question=A $object practices on the track. What is the displacement of the $object1 in meters, m, after he completes $laps@ qu.1.2.answer=0 m@ qu.1.2.random= $integer\=rint(4); $object\=switch($integer, "Nascar racer", "runner", "jogger", "bicyclist"); $object1\=switch($integer, "driver", "runner", "jogger", "bicyclist"); $laps\=switch(rint(22), "one lap?", "2 laps?", "3 laps?", "4 laps?", "5 laps?", "6 laps?", "7 laps?", "8 laps?", "9 laps?", "10 laps?", "11 laps?", "12 laps?", "13 laps?", "14 laps?", "15 laps?", "16 laps?", "17 laps?", "18 laps?", "19 laps?", "20 laps?", "21 laps?", "22 laps?");@ qu.1.2.hint.1=Displacement is the change in position. What is the change in position after finishing one lap? Consider the starting and ending points.@ qu.1.2.comment= 1. Displacement is the change in position. What is the change in position after finishing one lap? Consider the starting and ending points.

2. If the coordinate of the final position of a body is the same as the coordinate of its initial position, the displacement of the body equals to zero.@ qu.1.3.mode=Multiple Choice@ qu.1.3.editing=useHTML@ qu.1.3.name=Conceptual - Displacement@ qu.1.3.question=An object goes from one point in space to another. After it arrives at its destination, its displacement is_____________ the distance it traveled.@ qu.1.3.answer=4@ qu.1.3.choice.1=either greater than or equal to@ qu.1.3.choice.2=always greater than@ qu.1.3.choice.3=always equal to@ qu.1.3.choice.4=smaller than or equal to@ qu.1.3.choice.5=always smaller than@ qu.1.3.choice.6=either smaller or larger than@ qu.1.3.comment=

Curve 1 represents the total distance traveled from point A to point B and Curve 2 represents the displacement. @ qu.1.4.mode=Randomized Formula@ qu.1.4.editing=useHTML@ qu.1.4.name=Position vs. Time Graph@ qu.1.4.question=A body produces the position-versus-time graph shown below. Find the total displacement of the body in meters, m, on the interval from $interval.

@ qu.1.4.answer=$ans m@ qu.1.4.random= $integer\=rint(7); $interval\=switch($integer, "t_i = 0 s to t_f = 4 s", "t_i = 2 s to t_f = 6 s", "t_i = 4 s to t_f = 8 s", "t_i = 6 s to t_f = 10 s", "t_i = 2 s to t_f = 10 s", "t_i = 0 s to t_f = 8 s", "t_i = 0 s to t_f = 10 s"); $calculation\=switch($integer, -1000, -2000, 1000, 2000, 0, 0, 2000); $ans\=sig(1,$calculation);@ qu.1.4.hint.1=Displacement Dx = x_final - x_initial is the change in position.@ qu.1.4.comment=Displacement Dx = x_final - x_initial is the change in position.@ qu.2.topic=Ch 2, Sec 2 - Average Speed and Velocity@ qu.2.1.mode=Randomized Formula@ qu.2.1.editing=useHTML@ qu.2.1.name=Traveling Light@ qu.2.1.question=How long does it take the light to travel the distance of $distance x 10^$power m?@ qu.2.1.answer=$ans s (1 ? 0.1)@ qu.2.1.random= $distance\=rand(1,9.99,3); $power\=int(rint(70)+1); $ans\=sig(3, $distance*10^($power)/3E8);@ qu.2.2.mode=Multiple Choice@ qu.2.2.editing=useHTML@ qu.2.2.name=Negative Velocity (?)@ qu.2.2.question=An object's motion is described by the graph below. For which segment(s) is the velocity of the object negative?

@ qu.2.2.answer=2@ qu.2.2.choice.1=A and D@ qu.2.2.choice.2=C only@ qu.2.2.choice.3=B and D@ qu.2.2.choice.4=D only@ qu.2.2.choice.5=B only@ qu.2.2.hint.1=Notice that this is a position vs. time graph. Velocity is defined as displacement over time: v = Dx/Dt.@ qu.2.2.hint.2=Notice that the slope of a position graph is Dx/Dt = v.@ qu.2.2.comment= 1. Notice that this is a position vs. time graph. Velocity is defined as displacement over time: v = Dx/Dt.

2. The slope of a position graph is Dx/Dt = v. @ qu.2.3.mode=Randomized Formula@ qu.2.3.editing=useHTML@ qu.2.3.name=Bicyclist - Trip@ qu.2.3.question=A bicyclist traveling at $v1 mi/h covers $x1 mi. Then he reaches the downtown area, meets his friend, quickly delivers a book to him and starts moving toward a fast food place in the opposite direction at $v2 mi/h for another $x2 mi. How much time did the entire trip take?@ qu.2.3.answer= $ans s (1 ? 0.05);@ qu.2.3.random= $v1\=rand(12,17,3); $v2\=rand(10,11.99,3); $x1\=rand(5,9,3); $x2\=rand(2,4,3); $ans\=sig(3,($x1/$v1+$x2/$v2)*3600);@ qu.2.4.mode=Randomized Formula@ qu.2.4.editing=useHTML@ qu.2.4.name=Jogger@ qu.2.4.question=$name goes $activity every $timeofday. Usually he covers $distance km in approximately $time min. Find his average speed in m/s.@ qu.2.4.answer= $ans m/s (1 ? 0.05);@ qu.2.4.random= $name\=switch(rint(8), "John", "Travis", "Michael", "Joe", "Steven", "Mark", "George", "Fred"); $timeofday\=switch(rint(3), "afternoon", "evening", "morning"); $activity\=switch(rint(2), "running", "jogging"); $distance\=rand(3,5,3); $time\=rand(20,30,3); $ans\=sig(3,$distance*1000/($time*60));@ qu.2.4.hint.1=(Average Speed) = (Distance Traveled)/(Time Needed to Travel the Distance)@ qu.2.4.comment=(Average Speed) = (Distance Traveled)/(Time Needed to Travel the Distance)@ qu.2.5.mode=Randomized Formula@ qu.2.5.editing=useHTML@ qu.2.5.name=Soccer Ball@ qu.2.5.question=A soccer player kicks the ball giving it a speed of $speed km/h. If the ball travels in a straight line parallel to the ground, how long does it take the ball to reach another player $distance m away?@ qu.2.5.answer= $ans s (1 ? 0.05)@ qu.2.5.random= $speed\=rand(70,85,3); $distance\=rand(60,75,3); $ans\=sig(3,$distance/($speed/3.6))@ qu.2.6.mode=Randomized Formula@ qu.2.6.editing=useHTML@ qu.2.6.name=Truck@ qu.2.6.question=A truck loaded with hay travels on a straight road with a speed of $v1 mi/h for $x1 mi. It keeps going in the same direction for another $x2 mi at $v2 mi/h. Assuming that throughout the entire trip the truck was moving in the positive direction of the x axis, find the truck's average velocity on this trip.@ qu.2.6.answer= $ans m/s (1 ? 0.05)@ qu.2.6.random= $v1\=rand(60,70,3); $x1\=rand(30,40,3); $v2\=rand(45,55,3); $x2\=rand(15,25,3); $t1\=($x1/$v1)*3600; $t2\=($x2/$v2)*3600; $ans\=sig(3,(($x1+$x2)*1600)/($t1+$t2))@ qu.2.6.hint.1=(Average Velocity) = (Displacement)/(Time Elapsed)@ qu.2.6.comment= 1. (Average Velocity) = (Displacement)/(Time Elapsed)

2. Find total displacement and total time elapsed over the trip.
(Displacement)/(Velocity) = (Time Elapsed)
(Velocity) × (Time) = (Displacement)
@ qu.2.7.mode=Randomized Formula@ qu.2.7.editing=useHTML@ qu.2.7.name=Mouse vs. Two Cats@ qu.2.7.question=A narrow corridor is $length m long. Two cats begin to crawl toward each other from opposite ends of it with a speed of $speed m/s. At the same time small white mouse starts running away from one of the cats toward the other cat with the speed of $speed_mouse m/s. Once the mouse reaches the cat on the opposite end, it quickly turns around and moves in the opposite direction with the same speed that way trying to avoid both cats. What is the total distance that the small white mouse travels until the cats meet?@ qu.2.7.hint.1=The time that the mouse spends traveling is equal to the time it takes for the cats to meet.@ qu.2.7.answer= $ans m (1 ? 0.05)@ qu.2.7.random= $length\=rand(30,60,3); $speed\=rand(0.3,0.9,3); $speed_mouse\=rand(1.3,2.1,3); $ans\=sig(3,($length/(2*$speed))*$speed_mouse)@ qu.2.7.hint.2=(Speed) × (Time) = (Distance Traveled)@ qu.2.7.comment= 1. The time that the mouse spends traveling is equal to the time it takes for the cats to meet.

2. (Speed) × (Time) = (Distance Traveled) @ qu.2.8.mode=Randomized Formula@ qu.2.8.editing=useHTML@ qu.2.8.name=Round Trip@ qu.2.8.question=$name $action to another town. On the way to that town he travels with constant speed of $v1 mi/h. On the way back home he encounters a highway construction, so he had to reduce his speed to $v2 mi/h. Find $name$s average speed on this round trip?@ qu.2.8.answer= $ans m/s (1 ? 0.05)@ qu.2.8.random= $name\=switch(rint(7), "Tom", "John", "Peter", "Dave", "Mike", "Joseph", "Nathan"); $action\=switch(rint(3), "drives his car", "rides his motorcycle", "drives his truck"); $v1\=rand(50,65,3); $v2\=rand(35,40,3); $s\="'s"; $ans\=sig(3,(2*$v1*$v2/($v1+$v2))*16/36);@ qu.2.8.hint.1=The average of the two speeds will not equal the average speed of the entire trip, because he will spend more time traveling at a lower speed than at a higher speed.@ qu.2.8.hint.2=Suppose t₁ is the time it takes the driver to reach another town, t₂ is the time it takes him to drive back, d is the distance between the two towns, v₁ is the speed on the way to another town, and v₂ is the speed on the way back.

Thus,

@ qu.2.8.comment= 1. The average of the two speeds will not equal the average speed of the entire trip, because he will spend more time traveling at a lower speed than at a higher speed.

2. Suppose t₁ is the time it takes the driver to reach another town, t₂ is the time it takes him to drive back, d is the distance between the two towns, v₁ is the speed on the way to another town, and v₂ is the speed on the way back.

Thus,

@ qu.2.9.mode=Randomized Formula@ qu.2.9.editing=useHTML@ qu.2.9.name=Traveling Animal@ qu.2.9.question=A $animal_type travels at $v1 km/h for $x1 km, then it turns around and goes for $x2 km in the opposite direction with the speed of $v2 km/h. What is the average velocity of the $animal_type on this $total$km trip?@ qu.2.9.answer= $ans m/s (1 ? 0.05)@ qu.2.9.random= $animal_type=switch(rint(3), "dog", "rabbit", "cat"); $v1\=rand(3,6,3); $x1\=rand(1,2,3); $v2\=rand(6.3,12,3); $x2\=rand(0.4,0.9,3); $total\=sig(3,$x1+$x2); $km\="-km"; $displacement\=($x1-$x2)*1000; $time\=($x1/$v1+$x2/$v2)*3600; $ans\=sig(3,$displacement/$time);@ qu.2.9.hint.1=(Average Velocity) = (Displacement)/(Time Elapsed)@ qu.2.9.comment= 1. (Average Velocity) = (Displacement)/(Time Elapsed)

2. Because the animal changed direction, the displacement is less than the distance traveled over the entire trip.@ qu.3.topic=Ch 2, Sec 3 - Instantaneous Velocity@ qu.3.1.mode=Randomized Formula@ qu.3.1.editing=useHTML@ qu.3.1.name=Velocity Function - Linear@ qu.3.1.question=Velocity of a particle is represented by the function v(t) = $b - $a$time, where t is time in seconds. Find the instantaneous velocity of the particle at t = $time1 s.@ qu.3.1.answer= $ans m/s (1 ? 0.09)@ qu.3.1.random= $b1\=rint(98)+1; $b\=int($b1); $time="t"; $a\=rand(1,10,2); $time11\=rint(98)+1; $time1\=int($time11); $ans\=sig(2,$b-$a*$time1)@ qu.3.1.comment=Substitute the value of t in the v(t) equation in order to find the instantaneous velocity.@ qu.3.2.mode=Multiple Choice@ qu.3.2.editing=useHTML@ qu.3.2.name=Velocity Function - Parabolic@ qu.3.2.question=Velocity of a particle is given by the function v(t) = - t² + 5t + 1. The instantaneous velocity of that particle at t = 17 s is v (17) = - 203 m/s. Without doing any calculations, predict whether or not:@ qu.3.2.answer=3@ qu.3.2.choice.1=v(17) < v(10²⁰⁰²)@ qu.3.2.choice.2=v(17) = v(10²⁰⁰²)@ qu.3.2.choice.3=v(17) > v(10²⁰⁰²)@ qu.3.2.choice.4=v(10²⁰⁰²) = 0@ qu.3.2.choice.5=v(10²⁰⁰²) is not defined@ qu.3.2.comment=The graph of the function v(t) = - t² + 5t + 1 is an inverted parabola. So, as time goes on, the velocity becomes more negative. Note that the values t = 17 and t = 10²⁰⁰² are to the right of the vertex of the parabola.@ qu.3.3.mode=Randomized Formula@ qu.3.3.editing=useHTML@ qu.3.3.name=y(t) -> Average Velocity@ qu.3.3.question=The position of a body moving along the y axis is given by y = $b$t² - t + $a$t³. What is the body's average velocity on the interval between t = $t1 s and t = $t2 s? Give 3 significant figures.@ qu.3.3.answer= $ans m/s (1 ? 0.06)@ qu.3.3.random= $t\="t"; $b1\=rint(98)+1; $b\=int($b1); $a1\=rint(48)+1; $a\=int($a1); $t11\=rand(2,20,5); $t1\=int($t11); $t22\=rand(30,40,5); $t2\=int($t22); $displacement\=sig(3,($b*($t2)^2-$t2+$a*($t2)^3)-($b*($t1)^2-$t1+$a*($t1)^3)); $delta_t\=sig(3,$t2-$t1); $ans\=sig(3,$displacement/$delta_t)@ qu.3.3.hint.1=v_avg = Dx/Dt = (Change in Position)/(Change in Time) = (x₂ – x₁)/(t₂ – t₁).@ qu.3.3.comment= 1. v_avg = Dx/Dt = (Change in Position)/(Change in Time) = (x₂ – x₁)/(t₂ – t₁).

2. Evaluating the function y(t) at the initial and final times gives the initial and final positions. Plugging the values in v_avg = (x₂ – x₁)/(t₂ – t₁) gives average velocity.@ qu.3.4.mode=Randomized Formula@ qu.3.4.editing=useHTML@ qu.3.4.name=v(t)=?@ qu.3.4.question=The velocity of a body is represented by the graph below. What is v($t)$qmark
@ qu.3.4.answer=$ans m/s (1 ? 0.2)@ qu.3.4.random= $qmark\="?"; $integer\=rint(10); $t\=switch($integer, "0.5", "1", "1.5", "2", "2.5", "3", "3.5", "4", "4.5", "5"); $calculation\=switch($integer, 3.6, 2, 1.5, 1, 1, 1, 1, 1, 3, 5); $ans\=sig(2,$calculation)@ qu.3.4.comment=Find the velocity that corresponds to the time given.@ qu.4.topic=Ch 2, Sec 4 - Acceleration@ qu.4.1.mode=Randomized Formula@ qu.4.1.editing=useHTMl@ qu.4.1.name=v(t) - Average Acceleration@ qu.4.1.question=An object moves according to the graph below. Find the average acceleration of the object on the interval $interval$period
@ qu.4.1.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.1.random= $integer\=rint(5); $period\="."; $interval\=switch($integer, "A", "B", "C", "D", "E"); $calculation\=switch($integer,10,5,-10,2.5,0); $ans\=$calculation@ qu.4.1.hint.1=Acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.1.comment= 1. Acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).

2. The slope of the velocity vs. time graph is equal to acceleration: a = (v₂ – v₁)/(t₂ - t₁). @ qu.4.2.mode=Randomized Formula@ qu.4.2.editing=useHTML@ qu.4.2.name=v(t) => a(t)-?@ qu.4.2.question=A body produces the velocity-versus-time graph shown below. What is a($interval)?
@ qu.4.2.answer=$ans m/s^2 (1 ? 0.1)@ qu.4.2.random= $integer\=rint(4); $a\=rand(0.001,0.999,3); $b\=rand(1.01,1.99,3); $c\=rand(2.01,4.99,3); $d\=rand(5.01,5.99,3); $interval\=switch($integer,$a,$b,$c,$d); $calculation\=switch($integer,4,-3,-0.333,3); $ans\=sig(3,$calculation)@ qu.4.2.hint.1=Acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.2.comment= 1. Acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).

2. The slope of the velocity vs. time graph is equal to acceleration: a = (v₂ – v₁)/(t₂ - t₁). Because the slope of the graph is constant on intervals from t = 0 to t = 1, t = 1 to t = 2, t = 2 to t = 5, and t = 5 to t = 6, the acceleration within those intervals is constant. @ qu.4.3.mode=Randomized Formula@ qu.4.3.editing=useHTML@ qu.4.3.name=Magnitude of a@ qu.4.3.question=The velocity of a moving object (in SI units) is given by the following function: v(t) = $a11$t + $b. What is the magnitude of acceleration of this object?@ qu.4.3.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.3.random= $t\=t; $a11\=rand(-50,50,3); $a1\=($a11)^2; $b1\=rint(20)+1; $b\=int($b1); $ans\=sig(3,($a1)^(1/2))@ qu.4.3.comment=Acceleration is the slope of a velocity vs. time graph. The magnitude of acceleration is equal to the absolute value of the coefficient by t in the expression for v(t).@ qu.4.4.mode=Randomized Formula@ qu.4.4.editing=useHTML@ qu.4.4.name=Accelerating Car@ qu.4.4.question=A car reaches the speed of $speed mi/h in $time s. $question@ qu.4.4.answer=$ans $units (1 ? 0.05)@ qu.4.4.random= $time_later\=rand(1,5,3); $time\=rand(5,12,3); $integer\=rint(2); $question\=switch($integer,"What is the average acceleration of the car?","If the acceleration of the car remains constant, what is its speed after $time_later more seconds?"); $speed\=rand(45,85,3); $units\=switch($integer,"m/s^2","m/s"); $calculation\=switch($integer,($speed*1600/3600)/$time,(($speed*1600/3600)/$time)*($time+$time_later)); $ans\=sig(3,$calculation)@ qu.4.4.hint.1=Average acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.4.comment=Average acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.5.mode=Randomized Formula@ qu.4.5.editing=useHTML@ qu.4.5.name=Accelerating Insect@ qu.4.5.question=An insect had a speed of v_o = $v0 m/s at a certain time, and t = $time s later its speed was v₁ = $v1 m/s in the opposite direction. What was the magnitude of a_avg, the average acceleration of the insect during this $time s interval?@ qu.4.5.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.5.random= $v0\=rand(1,6,3); $v1=rand(4,7,3); $time\=rand(5,15,3); $ans\=sig(3,($v0+$v1)/$time)@ qu.4.5.hint.1=Average acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.5.comment=Average acceleration equals to the change in velocity divided by the change in time: a = Dv/Dt = (v₂ - v₁)/(t₂ - t₁).@ qu.4.6.mode=Randomized Formula@ qu.4.6.editing=useHTML@ qu.4.6.name=Constant Acceleration@ qu.4.6.question=The magnitude of constant acceleration of a body is $a m/s². What is its speed $t s later? Assume that the body starts moving from rest.@ qu.4.6.answer=$ans m/s (1 ? 0.05)@ qu.4.6.random= $a\=rand(1,5,3); $t\=rand(1,10,3); $ans\=sig(3,$a*$t)@ qu.4.6.hint.1.=(Acceleration) × (Time) = (Velocity)@ qu.4.6.comment=(Acceleration) × (Time) = (Velocity)@ qu.5.topic=Ch 2, Sec 5 - Motion With Uniform (Constant) Acceleration@ qu.5.1.mode=Randomized Formula@ qu.5.1.editing=useHTML@ qu.5.1.name=v(t) - Displacement@ qu.5.1.question=A body produces the velocity-versus-time graph shown below. Find the total displacement of the body $insert_phrase$period
@ qu.5.1.answer=$ans m (1 ? 0.1)@ qu.5.1.random= $period\="."; $integer\=rint(6); $insert_phrase\=switch($integer, "on the interval A", "on the interval B", "on the interval C", "on the intervals A and B", "on the intervals B and C", "on all the intervals"); $calculation\=switch($integer, 30, 150, 20, 180, 170, 200); $ans\=int($calculation)@ qu.5.1.hint.1=(Velocity) × (Time) = (Displacement)@ qu.5.1.comment= 1. (Velocity) × (Time) = (Displacement)

2. The area under a velocity vs. time curve gives displacement. Consider breaking this area into familiar geometrical figures (triangles and rectangles) in order to simplify the calculation.@ qu.5.2.mode=Randomized Formula@ qu.5.2.editing=useHTML@ qu.5.2.name=Fly Accelerating@ qu.5.2.question=A fly has a constant acceleration of +$a m/s². At one point in time its velocity is $v m/s. What is the velocity of the fly $time s earlier?@ qu.5.2.answer= $ans m/s (1 ? 0.05)@ qu.5.2.random= $a\=rand(1,3,3); $v\=rand(-3,3,3); $time\=rand(.1,2,3); $ans\=sig(3, $v+$a*(-$time))@ qu.5.2.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.2.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: a, v, t
Unknown variables: v₀
Variables not involved: x

Solving the equation v = v₀ + at for the unknown variable, we obtain v₀ = v – at.
Plugging in respective numerical values, calculate v₀ in m/s. @ qu.5.3.mode=Randomized Formula@ qu.5.3.editing=useHTML@ qu.5.3.name=Animal Running@ qu.5.3.question=A $animal starts running from rest with constant acceleration of $a m/s². Assuming that its speed becomes constant after $ta s, find how far the $animal travels between t = 0 s and t = $t s.@ qu.5.3.answer=$ans m (1 ? 0.05)@ qu.5.3.random= $animal\=switch(rint(3),"rabbit","dog","fox"); $a\=rand(0.5,2,3); $ta\=rand(1,4,3); $t\=rand(4,9,3); $ans\=sig(3,0.5*$a*($ta)^2+($a*$ta)*($t-$ta))@ qu.5.3.hint.1=When the speed becomes constant, the acceleration becomes zero. Thus, this problem has two parts: one in which you need to account for acceleration and one in which the acceleration is zero.@ qu.5.3.comment= When the speed becomes constant, the acceleration becomes zero. Thus, this problem has two parts: one in which you need to account for acceleration and one in which the acceleration is zero.

Part 1 (Constant Acceleration)

Known variables: a, v₀, t
Unknown variables: x
Variables not involved: v

Part 2 (Zero Acceleration)

Known variables: a, v₀, t
Unknown variables: x
Variables not involved: v

While the animal is moving with constant acceleration, its motion is described by the following equation: x = v₀t + (¹/₂)at². Calculate x and find how far the animal traveled while it was accelerating. Use the equation v = v₀ + at to calculate the speed of the animal immediately before its speed becomes constant. Use this value of speed in the equation x = v₀t + (¹/₂)at² to calculate how far the animal traveled at constant speed. @ qu.5.4.mode=Randomized Formula@ qu.5.4.editing=useHTML@ qu.5.4.name=Airplane Accelerating@ qu.5.4.question=An airplane has an initial velocity of v₀ = $v0 m/s when it starts to accelerate. Find the final velocity of the airplane after $t s if its acceleration is $a m/s².@ qu.5.4.answer=$ans m/s (1 ? 0.05)@ qu.5.4.random= $v0\=rand(190,250,3); $t\=rand(1.5,5,3); $a\=rand(8,15,3); $ans\=sig(3,$v0+$a*$t)@ qu.5.4.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.4.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: a, v₀, t
Unknown variables: v
Variables not involved: x
Use the equation v = v₀ + at to find the final velocity of the airplane in m/s. @ qu.5.5.mode=Randomized Formula@ qu.5.5.editing=useHTML@ qu.5.5.name=Crossing the Bridge@ qu.5.5.question=A $vehicle traveling with a constant acceleration of $a m/s² crosses a $l-m-long bridge in $t s. What was the velocity at the beginning of the bridge?@ qu.5.5.answer=$ans m/s (1 ? 0.05)@ qu.5.5.random= $vehicle\=switch(rint(2),"car","motorcycle"); $a\=rand(0.5,5,3); $l\=rand(300,600,3); $t\=rand(8,10,3); $ans\=sig(3,$l/$t-0.5*$a*$t)@ qu.5.5.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.5.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: a, x, t
Unknown variables: v₀
Variables not involved: v
Solving x = v₀t + (¹/₂)at² for v₀, we get v₀ = (x – (¹/₂)at²)/t. Substitute respective numerical values and calculate v₀ in m/s. @ qu.5.6.mode=Randomized Formula@ qu.5.6.editing=useHTML@ qu.5.6.name=Bullet@ qu.5.6.question=A bullet traveling with the speed of $v m/s hits a brick wall. Find the magnitude of deceleration of the bullet if it travels $d cm inside the brick wall.@ qu.5.6.answer=$ans m/s^2 (1 ? 0.05)@ qu.5.6.random= $v\=rand(700,900,3); $d\=rand(2,8,3); $ans\=sig(3,100*($v)^2/(2*$d))@ qu.5.6.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.6.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: v, x, v₀
Unknown variables: a
Variables not involved: t
Solving v² = v₀² + 2ax, we obtain a = (v² - v₀²)/(2x). Substitute respective numerical values and calculate a in m/s². @ qu.5.7.mode=Randomized Formula@ qu.5.7.editing=useHTML@ qu.5.7.name=Train - Deceleration@ qu.5.7.question=A traveling train has a velocity of v₀ = $v km/h and is l = $l m away from a large obstacle when the machinist slams on the brakes. The train stops just before the obstacle barely avoiding the collision. What is the magnitude of the train's deceleration a?@ qu.5.7.answer=$ans m/s^2 (1 ? 0.05)@ qu.5.7.random= $v\=rand(90,120,3); $l\=rand(121,200,3); $ans\=sig(3,($v/3.6)^2/(2*$l))@ qu.5.7.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.7.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: v, x, v₀
Unknown variables: a
Variables not involved: t
Solving v² = v₀² + 2ax, we obtain a = (v² - v₀²)/(2x). Substitute respective numerical values and calculate a in m/s².@ qu.5.8.mode=Randomized Formula@ qu.5.8.editing=useHTML@ qu.5.8.name=Rolling Ball@ qu.5.8.question=A ball starts rolling down a very long incline with an acceleration of $a m/s². How far does the ball travel in $t s?@ qu.5.8.answer=$ans m (1 ? 0.05)@ qu.5.8.random= $a\=rand(0.2,7.6,3); $t\=rand(5,30,3); $ans\=sig(3,0.5*$a*($t)^2)@ qu.5.8.hint.1=Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.5.8.comment= 1. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: a, t, v₀
Unknown variables: x
Variables not involved: v
Use x = v₀t + (¹/₂)at² to calculate x in meters, m.@ qu.5.9.mode=Randomized Formula@ qu.5.9.editing=useHTML@ qu.5.9.name=Decelerating Automobile@ qu.5.9.question=A $vehicle is traveling on a highway with a speed of $v mi/h on a foggy day. Suddenly the driver sees another $vehicle traveling at $v1 mi/h just $x m away directly ahead of him. What must me be the magnitude of minimum deceleration of the faster $vehicle in order for the collision to be barely avoided?@ qu.5.9.answer=$ans m/s^2 (1 ? 0.05)@ qu.5.9.random= $vehicle\=switch(rint(3),"vehicle","car","truck"); $v\=rand(55,65,3); $v1\=rand(25,35,3); $x\=rand(7,18,3); $ans\=sig(3,(($v-$v1)*16/36)^2/(2*$x))@ qu.5.9.hint.1=In order for the collision to be avoided, the faster vehicle must decelerate to the speed of the other vehicle. In order for the magnitude of deceleration to be minimum, the deceleration must occur over the entire distance between the two vehicles.@ qu.5.9.comment= 1. In order for the collision to be avoided, the faster vehicle must decelerate to the speed of the other vehicle. In order for the magnitude of deceleration to be minimum, the deceleration must occur over the entire distance between the two vehicles.

2. Suppose that v₁ is the speed of the slower vehicle and v₂ is the speed of the faster vehicle. To approach the solution to this problem, assume that the slower vehicle is stopped (not moving). Thus, relative to the slower vehicle, the faster vehicle is moving with the speed v₀ = v₂ – v₁.
Consider the motion of the faster vehicle as it decelerates from speed v₀ = v₂ – v₁ to zero over the initial distance between the two vehicles.
Use the equation v² = v₀² + 2ax to solve for the minimum acceleration necessary to avoid collision, where v is the final velocity of the faster vehicle and is equal to zero (assuming the slower vehicle is considered stopped).@ qu.5.10.mode=Randomized Formula@ qu.5.10.editing=useHTML@ qu.5.10.name=Speeding Motorist@ qu.5.10.question=A speeding motorist passes a stopped police car. At the moment the motorist passes, the police car starts from rest with a constant acceleration of $a m/s². The speeding motorist continues with constant velocity until caught by the police car $t s later. How fast was the speeding car going?@ qu.5.10.answer=$ans m/s (1 ? 0.05)@ qu.5.10.random= $a\=rand(2,5,3); $t\=rand(10,18,3); $ans\=sig(3,0.5*$a*$t)@ qu.5.10.hint.1=At the point where the police car catches up with the driver, the position x of the two is the same.@ qu.5.10.comment= 1. At the point where the police car catches up with the driver, the position x of the two is the same.

2. Using x = v₀t + (¹/₂)at², we set x_{police car} = x_{speeding driver}.
So, x_{police car} = v_{0 police car} t + (¹/₂)a_{police car} t² and x_{speeding driver} = v_{speeding driver} t because the speeding car has constant velocity. Since the police car started moving from rest, we obtain:
(¹/₂)a_{police car} t² = v_{speeding driver} t.
Now we can solve for the speed of the speeding driver.@ qu.5.11.mode=Randomized Formula@ qu.5.11.editing=useHTML@ qu.5.11.name=A->C Car Trip@ qu.5.11.question=A car located at point A starts accelerating from rest with an acceleration of $a km/h². When it reaches point B ($l m away), the driver shifts into neutral gear and the car gradually decelerates at $a1 km/h² and stops at point C. How much time did it take the car to travel from point A to point C?@ qu.5.11.answer=$ans s (1 ? 0.05)@ qu.5.11.random= $a\=rand(3,9,3); $a1\=rand(0.2,1,3); $l\=rand(30,70,3); $ans\=sig(3,((2*$l*$a)^(1/2)/$a1)*3600)@ qu.5.11.hint.1=Consider the intervals from A to B and from B to C separately.@ qu.5.11.comment= 1. Consider the intervals from A to B and from B to C separately.

2. Interval from A to B:
Known variables: a, x, v₀
Unknown variables: t
Variables not involved: v
Use the equation x = v₀t + (¹/₂)at² to find the time the car spends accelerating on the interval from A to B. Then use this value of time to determine the final velocity of the car on the interval from A to B and use this velocity as an initial velocity for the interval from B to C.

Interval from B to C:
Known variables: a, v (equals to zero because the car comes to a stop), v₀ (equals to the final velocity on the interval from A to B)
Unknown variables: t
Variables not involved: x

Use v = v₀ + at find the time it takes the car to stop.

To obtain the final answer, add the two values of time to find the total time it takes the car to travel from point A to point C.@ qu.6.topic=Ch 2, Sec 6 - Free-Falling Bodies@ qu.6.1.mode=Multiple Choice@ qu.6.1.editing=useHTML@ qu.6.1.name=Ball Thrown Straight Up@ qu.6.1.question=You are throwing a ball straight up in the air. At the highest point, the ball's...@ qu.6.1.answer=3@ qu.6.1.choice.1=velocity and acceleration are zero@ qu.6.1.choice.2=velocity is nonzero but its acceleration is zero@ qu.6.1.choice.3=acceleration is nonzero, but its velocity is zero@ qu.6.1.choice.4=velocity and acceleration are both nonzero@ qu.6.1.comment=The ball moves with the acceleration due to gravity all throughout the flight, but because the direction of acceleration is opposite to the direction of the initial velocity of the ball (the acceleration due to gravity is directed radially toward the center of Earth), the acceleration causes velocity to reach 0 m/s and change direction at the highest point.@ qu.6.2.mode=Multiple Choice@ qu.6.2.editing=useHTML@ qu.6.2.name=Two Balls@ qu.6.2.question=A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed. Neglecting the air resistance, the ball to his the ground below the cliff with the greater speed is the one initially thrown@ qu.6.2.answer=3@ qu.6.2.choice.1=upward@ qu.6.2.choice.2=downward@ qu.6.2.choice.3=neither - they both hit the ground at the same time@ qu.6.2.comment=The position vs. time graph generated by the ball thrown straight up is shown below.

Suppose that the line y = 0 is the level of the cliff. According to the graph, as the ball comes down from its highest point and crosses the y = 0 line, it reaches the speed equal to its initial speed. At this point the ball will repeat the motion of the ball thrown downward. Thus, the two balls hit the ground at the same speed though at different points in time.@ qu.6.3.mode=Multiple Choice@ qu.6.3.editing=useHTML@ qu.6.3.name=Acceleration Due To Gravity@ qu.6.3.question=If you drop and object in the absence of air resistance, it accelerates downward at 9.8 m/s². If instead you throw it downward, its downward acceleration after release is:@ qu.6.3.answer=2@ qu.6.3.choice.1=less than 9.8 m/s²@ qu.6.3.choice.2=9.8 m/s²@ qu.6.3.choice.3=greater than 9.8 m/s²@ qu.6.3.choice.4=zero@ qu.6.3.comment=If an object is thrown then after it is released, there is no force other than gravitational force acting on it. Thus, the object moves with the acceleration due to gravity.@ qu.6.4.mode=Multiple Choice@ qu.6.4.editing=useHTML@ qu.6.4.name=Acceleration Due To Gravity (2)@ qu.6.4.question=As a freely falling object speeds up, what happens to its acceleration due to gravity?@ qu.6.4.answer=2@ qu.6.4.choice.1=The acceleration due to gravity is zero@ qu.6.4.choice.2=The acceleration due to gravity stays the same@ qu.6.4.choice.3=The acceleration due to gravity decreases@ qu.6.4.choice.4=The acceleration due to gravity increases@ qu.6.4.comment=Acceleration due to gravity is always constant. Only the velocity changes.@ qu.6.5.mode=Randomized Formula@ qu.6.5.editing=useHTML@ qu.6.5.name=Thrown Spherical Object@ qu.6.5.question=If a spherical object is thrown straight up into the air from the ground level with the velocity of $v m/s, what is the maximum height that it will reach?@ qu.6.5.answer=$ans m (1 ? 0.05)@ qu.6.5.random= $v\=rand(3,35,3); $ans\=sig(3,($v)^2/(2*9.81))@ qu.6.5.hint.1=Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.6.5.comment= 1. Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: v₀, v, a
Unknown variables: y
Variables not involved: t

Solve the equation v² = v₀² + 2ay for y to find the maximum height.@ qu.6.6.mode=Randomized Formula@ qu.6.6.editing=useHTML@ qu.6.6.name=Falling Flowerpot@ qu.6.6.question=A flowerpot accidentally falls straight down off the window that is $h ft above the ground. How long does it take the pot to reach the base of the building?@ qu.6.6.answer=$ans s (1 ? 0.1)@ qu.6.6.random= $h\=rand(15,35,3); $ans\=sig(3,(2*$h*0.3048/9.81)^(1/2))@ qu.6.6.hint.1=Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.6.6.comment= 1. Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: y, v₀, a
Unknown variables: t
Variables not involved: v

Solve y = y₀ + v₀t + (¹/₂)at² for t to obtain the answer. @ qu.6.7.mode=Randomized Formula@ qu.6.7.editing=useHTML@ qu.6.7.name=Snowball Thrown Downward@ qu.6.7.question=A snowball is thrown downward with some initial velocity v₀ from a height of $h m. Knowing that it takes the ball $t s to reach the ground, find v₀.@ qu.6.7.answer=$ans m/s (1 ? 0.05)@ qu.6.7.random= $h\=rand(20,30,3); $t\=rand(0.6,1.3,3); $ans\=sig(3,$h/$t-0.5*9.81*$t)@ qu.6.7.hint.1=Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use. @ qu.6.7.comment= 1. Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: y, t, a
Unknown variables: v₀
Variables not involved: v

Solve y = y₀ + v₀t + (¹/₂)at² for v₀ to obtain the answer.@ qu.6.8.mode=Randomized Formula@ qu.6.8.editing=useHTML@ qu.6.8.name=Large Rock (Falling)@ qu.6.8.question=A large rock falls straight down off a cliff. An observer at the base of the cliff notices that the rock reaches the ground in about $t s. What is the height of the cliff?@ qu.6.8.answer=$ans m (1 ? 0.05)@ qu.6.8.random= $t\=rand(3,15,3); $ans\=sig(3,0.5*9.81*($t)^2)@ qu.6.8.hint.1=Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.@ qu.6.8.comment= 1. Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: v₀, t, a
Unknown variables: y
Variables not involved: v

Calculate y using y = y₀ + v₀t + (¹/₂)at². @ qu.6.9.mode=Randomized Formula@ qu.6.9.editing=useHTML@ qu.6.9.name=Sandbag (Falling)@ qu.6.9.question=A helicopter is ascending with the speed of $v km/h. When it reaches the height of $h m, a sandbag is released from the helicopter. How long does it take the sandbag to reach the ground?@ qu.6.9.answer=$ans s (1 ? 0.05)@ qu.6.9.random= $v\=rand(8,15,3); $h\=rand(100,300,3); $v1\=$v/3.6; $ans\=sig(3,(($v1+(($v1)^2+19.62*$h)^(1/2))/(9.81)))@ qu.6.9.hint.1=Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use. @ qu.6.9.comment= 1. Motion with the acceleration due to gravity is a case of motion with constant acceleration. Determine the variables that are known, unknown, and those that are not involved. Using this information, determine which kinematics equation to use.

2.
Known variables: y, a, v₀
Unknown variables: t
Variables not involved: v

Solve the equation y = y₀ + v₀t + (¹/₂)at² for t. Note that this is a quadratic equation that has two real solutions. Choose the solution that has the most physical meaning. @