One-Dimensional Kinematics Review

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Part E: Computational Problems


43. Determine the acceleration (in m/s2) of an object which ... .

  1. moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds
  2. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds
  3. changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds
  4. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m

Answer: See answers, explanations and calculations below.

a. If the speed and direction of an object is constant, then the acceleration is 0 m/s2.

b. The acceleration is the velocity change per time ratio:

a = (Velocity Change)/t = (23.5 m/s - 12.1 m/s) / (7.81 s) = 1.46 m/s2.

c. The acceleration is the velocity change per time ratio:

a = (Velocity Change)/t = (60.0 mi/hr - 0.0 mi/hr) / (4.20 s) = 14.3 mi/hr/s.

14.3 mi/hr/s * (1.0 m/s) / (2.24 mi/hr) = 6.38 m/s2.

d. The acceleration value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 33.4 m/s; vf = 18.9 m/s; and d = 109 m. Using the equation vf2 = vo2 + 2*a*d, the acceleration can be computed.

a = (vf2 - vo2) / (2*d) = [(18.9 m/s)2 - (33.4 m/s)2 ] / (2 * 109 m) = -3.48 m/s2.
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44. Determine the magnitude of the displacement (in meters) of an object which ... .

  1. moves from Hither to Yon (with an average speed of 28.0 m/s) and then back to Hither (with an average speed of 28.0 m/s) if both the forward and the return trip take 46 minutes each.
  2. moves at a constant speed of 8.30 m/s in a straight line for 15.0 seconds.
  3. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s
  4. accelerates from rest at a rate of 3.67 m/s2 for 12.1 seconds
  5. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s2 for 17.0 seconds

Answer: See answers, explanations and calculations below.

a. Since this is a round-trip journey, the overall displacement is 0 m.

b. Since the velocity is constant, the displacement can be found by multiplying the velocity by the time.

d = v*t = (8.30 m/s) * (15.0 s) = 125 m

c. A displacement value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 38.1 m/s; vf = 17.6 m/s; and a = -4.35 m/s/s. Using the equation vf2 = vo2 + 2*a*d, the displacement can be computed.

d = (vf2 - vo2) / (2*a) = [(17.6 m/s)2 - (38.1 m/s)2 ] / (2 * -4.35 m/s/s) = 131 m.

d. A displacement value can be calculated using other kinematic equations if a different set of kinematic quantities is known. Here we know that: vo = 0.0 m/s; t = 12.1 s; and a = 3.67 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.

d = (0 m/s)*(12.1 s) + 0.5*(3.67 m/s/s)*(12.1 s)2 = 269 m.

e. Here the displacement value is calculated using the same kinematic equation. We know that: vo = 12.2 m/s; t = 17.0 s; and a = 1.88 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.

d = (12.2 m/s)*(17.0 s) + 0.5*(1.88 m/s/s)*(17.0 s)2 = 479 m.
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45. The hare is sleeping at a location that is 1200 m from the finish line. The tortoise passes him at a steady speed of 5.0 cm/s. If the hare finally wakes up 6.5 hours later, then what minimum acceleration (assumed constant) must he have in order to pass the tortoise before the finish line.

Answer: 0.0067 m/s2

Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).

The tortoise, moving at a constant speed, will cover the 1200 m in a time of:

ttortoise = d/vtortoise = (1200 m) / (0.050 m/s) = 24000 s = 6.666... hrs

The hare will sleep for 6.5 hrs (23400 s) before starting, and so will have only 0.1666... hrs (600 s) to accelerate to the finish line. So the acceleration of the hare can be determine using a kinematic equation. The know information about the hare's motion is: t = 600 s; d = 1200 m; vo = 0 m/s. The best equation is d = vo* t + 0.5*a*t2. The vo* t term cancels and the equation can be algebraically rearranged and solved for a:

a = 2*d / t2 = 2 * (1200 m) / (600 s)2 = 0.0067 m/s2.
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46. A Gold Car moving at 12.0 m/s passes a Green Car while the Green Car is at rest at a stoplight. The Green Car immediately accelerates at a rate of +1.80 m/s/s for 11.0 seconds seconds and then maintains a constant speed. After how much time (relative to the initial starting time) must the Green Car drive before catching up with the Gold Car

Answer: 14.0 s

(As mentioned in the previous problem ...) Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).

Here the gold car travels with a constant speed for a time of t seconds (where t is the total time of travel for both cars). The distance traveled by the gold car is given by the kinematic equation d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as

d = vo* t + 0.5*a*t2 = (12.0 m/s)*t, or

dgold = 12.0* t

For the green car, there is an accelerated period and then a constant speed period. The distance traveled during the accelerated period (d1green) is found from the same kinematic equation. For the green car, the first term cancels and the distance is

d1green = vo* t + 0.5*a*t2 = 0.5*(1.80 m/s2)*(11.0 s)2, or

d1green = 108.9 m

Once the green car has accelerated for 11 seconds, it maintains a constant speed for the remaining time, given by the expression t - 11 s. The speed at which the green car is traveling during this time can be computed using the equation:

vfgreen = vo + a*t = (1.80 m/s2) *(11.0 s) = 19.8 m/s

The distance traveled by the green car during this constant speed portion of its motion (d2green) can be computed using the kinematic equation. d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as

d2green = vo* t + 0.5*a*t2 = (19.8 m/s) * (t - 11 s) = 19.8*t - 217.8, or

d2green = 19.8*t - 217.8

So the total distance traveled by the green car is given by the expression:

dgreen = d1green + d2green = 108.9 + 19.8*t - 217.8

dgreen = 19.8*t - 108.9

When the green car catches up to the gold car their distance traveled will be the same. So the time t can be determined by setting the two expressions for distance equal to each other and solving for t.

12.0* t = 19.8*t - 108.9

108.9 = 7.80*t

t = (108.9) / (7.80)

t = 13.96 s = 14.0 s

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47. Ima Rilla Saari is cruising at 28.0 m/s down Lake Avenue and through the forest preserve. She notices a deer jump into the road at a location 62.0 m in front of her. Ima first reacts to the event, then slams on her brakes and decelerates at -8.10 m/s2, and ultimately stops a picometer in front of the frozen deer. What is Ima's reaction time? (i.e., how long did it take Ima to react to the event prior to decelerating?)

Answer: 0.486 s

Ima's total distance traveled (62.0 m) can be broken into two segments - a reaction distance (drxn) and a braking distance (dbraking). The reaction distance is the distance Ima moves prior to braking; she will move at constant speed during this time of trxn. The braking distance is the distance which Ima moves when her foot is on the brake and she decelerates from 28.0 m/s to 0.0 m/s. The braking distance can be computed first using the following kinematic equation: vf2 = vo2 + 2*a*d. The known information for this braking period is: vo = 28 m/s; vf = 0 m/s; and a = -8.10 m/s/s. The substitutions and solution are shown below.

dbraking = (vf2 - vo2) / (2*a) = [(0 m/s)2 - (28.0 m/s)2 ] / (2 * -8.10 m/s/s) = 48.40 m.

Since Ima's car requires 48.40 m to brake, she can travel a maximum of 13.6 m during the reaction period. The relationship between reaction time, speed and reaction distance is given by the equation

drxn = v * trxn

Substituting 13.6 m for drxn and 28.0 m/s for v, the reaction time can be computed:

trxn = (13.6 m) / (28.0 m/s) = 0.486 s
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48. A two-stage rocket accelerates from rest at +3.57 m/s/s for 6.82 seconds. It then accelerates at +2.98 m/s/s for another 5.90 seconds. After the second stage, it enters into a state of free fall. Determine:

  1. the maximum speed
  2. the maximum altitude
  3. the height of the rocket after 20.0 seconds
  4. the total time the rocket is in the air (assuming it is launched from the ground)

Answer: See answers and explanations below.

This problem can be approached by either the use of a velocity-time graph or the use of kinematic equations (or a combination of each). Whatever the approach, it is imperative to break the multistage motion up into its three different acceleration periods. The use of kinematic equations is only appropriate for constant acceleration periods. For this reason, the complex motion must be broken up into time periods during which the acceleration is constant. These three time periods can be seen on the velocity-time graph by three lines of distinctly different slope. The diagram at the right provides a depiction of the motion; strategic points are labeled. These points will be referred to in the solutions below. The velocity-time plot below will be used throughout the solution; note that the same strategic points are labeled on the plot.

a. The maximum speed occurs after the second stage or acceleration period (point C). After this time, the upward-moving rocket begins to slow down as gravity becomes the sole force acting upon it. To determine this speed (vc), the kinematic equation vf = vo + a*t will be used twice - once for each acceleration period.

First Stage: vB = vA + a*t = 0 m/s + (3.57 m/s/s) * (6.82 s) = 24.3 m/s

Second Stage: vC = vB + a*t = 24.3 m/s + (2.98 m/s/s) * (5.90 s) = 41.9 m/s

b. The maximum altitude occurs at point D, sometime after the second stage has ceased and the rocket finally runs out of steam. The velocity at this point is 0 m/s (it is at the peak of its trajectory). The altitude at this point is the cumulative distance traveled from t = 0 s to t = tD. This distance is the distance for the first stage, the second stage and the deceleration period (C to D). These distances correspond to the area on the v-t graph; they are labeled A1, A2, and A3 on the graph. They are calculated and summed below.

A1 = 0.5*b*h = 0.5 * (24.3 m/s) * (6.82 s) = 82.86 m

A2 = b*h + 0.5*b*h (a triangle on top of a square)

A2 = (24.3 m/s) * (5.9 s) + 0.5 * (41.9 m/s - 24.3 m/s) * (5.9 s) = 196.65 m

It will be necessary to know the time from point C to point D in order to determine A3. This time can be determined using the kinematic equation vf = vo + a*t for which vf = 0 m/s and vo = 41.9 m/s and a = -9.8 m/s/s.

vD = vC + a*t

0 m/s = 41.9 m/s + (-9.8 m/s/s) * t

t = 4.28 s

Now A3 can be determined using the v-t graph. The area is a triangle and is calculated as

A3 = 0.5*b*h = 0.5 * (41.9 m/s) * (4.82 s) = 89.57 m

The maximum altitude is the sum of the three distances (areas)

Max. altitude = 82.86 m + 196.65 m + 89.57 m = 369 m

c. When the rocket reaches point D, the time is 17.0 seconds. The altitude at 20.0 seconds will be the 369 meters risen above the launch pad from point A to point D minus the distance fallen from the peak from 17.0 to 20.0 seconds. This distance would be represented by a negative area on the velocity-time graph. The area is a triangle and can be computed if the velocity at 20 seconds is known. It can be calculated using a kinematic equation and then used to determine the area of a triangle. Alternatively, a kinematic equation can be used to determine the distance fallen during these 3.0 seconds. The work is shown below:

d = vo* t + 0.5*a*t2 = 0.5 * (-9.8 m/s/s) * (3.0 s)2 = 44.1 m

The altitude at 20 seconds is therefore the ~369 m risen in the first 17 seconds minus the ~44 m fallen in the next 3 seconds. The answer is 325 m.

d. The rocket rises 369 m in the first 17.0 seconds. In the time subsequent of this, the rocket must fall 369 meters. The time to fall 369 m can be found from the same kinematic equation used in part c.

d = vo* t + 0.5*a*t2

-369 m = 0.5 * (-9.8 m/s/s) * t2

t = 8.68 seconds

This time can be added onto the 17.0 seconds to determine the time at which the rocket lands: 25.7 seconds.

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49. In a 200.0-m relay race (each leg of the race is 50.0 m long), one swimmer has a 0.450 second lead and is swimming at a constant speed of 3.90 m/s towards the opposite end of the pool. What minimum speed must the second swimmer have in order to catch up with the first swimmer by the end of the pool?

Answer: 4.04 m/s

Both swimmers swim the same distance (50 m) at constant speed. Swimmer A (who was just arbitrarily named) gets a 0.450 second head start. So swimmer B must travel faster in order to finish the race in less time than swimmer A. First, the time required of swimmer A to complete the 50l0 m at 3.90 m/s can be computed. The time is

tA = d/vA = (50.0 m) / (3.90 m/s) = 12.82 s

Thus, swimmer B must finish the same 50.0 m in 12.37 seconds (12.82s - 0.45 s). So the speed of swimmer B can be computed as

vB = d/tB = (50.0 m) / (12.37 s) = 4.04 m/s
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50. A drag racer accelerates from rest at an average rate of +13.2 m/s2 for a distance of 100. m. The driver coasts for 0.500 seconds and then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180. m, what minimum deceleration rate must the racer have in order to stop prior to the end of the track?

Answer: -24 m/s/s

This problem can be approached by first determining the distance over which the dragster decelerates. This distance will be less than 80. meters by an amount equal to the distance which the dragster coasts after crossing the finish line. See diagram.

The distance traveled by the dragster prior to braking is 100 m plus the coasting distance. The coasting distance can be determined if the speed of the dragster at the end of 100 m is determined. So first, a kinematic equation will be used to determine the speed and then the coasting distance will be computed.

Using the equation vf2 = vo2 + 2*a*d, the speed after 100 m can be determined. This substitution and solution is shown below.

vf2 = vo2 + 2*a*d = 2*(13.2 m/s/s)*(100. m) = 2640 m2/s2

vf = 51.4 m/s

Coasting at 51.38 m/s for 0.500 s will lead to a distance traveled of 25.7 m.

Once the coasting period is over, there is a short distance left to decelerate to a stop. This distance is

180. m - 100. m - 25.7 m = 54 m

Now the same kinematic equation can be used to determine the deceleration rate during the last 54 m of the track. Known information is: vo = 51.4 m/s; vf = 0 m/s; and d = 54 m. Using the equation vf2 = vo2 + 2*a*d, the acceleration can be computed.

a = (vf2 - vo2) / (2*d) = = [(51.4 m/s)2 - (0 m/s)2 ] / (2 * 54 m) = -24 m/s2.
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