Sample Problems and Solutions
Earlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented. These problems allow any student of physics to test their understanding of the use of the four kinematic equations to solve problems involving the one-dimensional motion of objects. You are encouraged to read each problem and practice the use of the strategy in the solution of the problem. Then click the button to check the answer or use the link to view the solution.
Check Your Understanding
- An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
- A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
- Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
- A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
- A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
- Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?
- A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
- An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?
- A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).
- A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
- If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
- A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
- A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)
- The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
- A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)
- A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
- It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.
- A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
- A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.
- With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
Solutions to Above Problems
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Given:
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a = +3.2 m/s2
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t = 32.8 s
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vi = 0 m/s
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Find:
d = ??
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d = vi*t + 0.5*a*t2
d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2
d = 1720 m
Return to Problem 1
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Given:
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d = 110 m
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t = 5.21 s
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vi = 0 m/s
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Find:
a = ??
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d = vi*t + 0.5*a*t2
110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2
110 m = (13.57 s2)*a
a = (110 m)/(13.57 s2)
a = 8.10 m/ s2
Return to Problem 2
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Given:
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a = -9.8 m
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t = 2.6 s
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vi = 0 m/s
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Find:
d = ??
vf = ??
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d = vi*t + 0.5*a*t2
d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s2)*(2.60 s)2
d = -33.1 m (- indicates direction)
vf = vi + a*t
vf = 0 + (-9.8 m/s2)*(2.60 s)
vf = -25.5 m/s (- indicates direction)
Return to Problem 3
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Given:
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vi = 18.5 m/s
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vf = 46.1 m/s
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t = 2.47 s
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Find:
d = ??
a = ??
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a = (Delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2
d = 45.7 m + 34.1 m
d = 79.8 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Return to Problem 4
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Given:
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vi = 0 m/s
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d = -1.40 m
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a = -1.67 m/s2
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Find:
t = ??
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d = vi*t + 0.5*a*t2
-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2
-1.40 m = 0+ (-0.835 m/s2)*(t)2
(-1.40 m)/(-0.835 m/s2) = t2
1.68 s2 = t2
t = 1.29 s
Return to Problem 5
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Given:
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vi = 0 m/s
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vf = 444 m/s
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t = 1.83 s
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Find:
a = ??
d = ??
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a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)
a = 243 m/s2
d = vi*t + 0.5*a*t2
d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2
d = 0 m + 406 m
d = 406 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Return to Problem 6
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Given:
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vi = 0 m/s
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vf = 7.10 m/s
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d = 35.4 m
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Find:
a = ??
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vf2 = vi2 + 2*a*d
(7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)
50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a
(50.4 m2/s2)/(70.8 m) = a
a = 0.712 m/s2
Return to Problem 7
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Given:
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vi = 0 m/s
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vf = 65 m/s
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a = 3 m/s2
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Find:
d = ??
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vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m2/s2)/(6 m/s2) = d
d = 704 m
Return to Problem 8
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Given:
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vi = 22.4 m/s
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vf = 0 m/s
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t = 2.55 s
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Find:
d = ??
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d = (vi + vf)/2 *t
d = (22.4 m/s + 0 m/s)/2 *2.55 s
d = (11.2 m/s)*2.55 s
d = 28.6 m
Return to Problem 9
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Given:
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a = -9.8 m/s2
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vf = 0 m/s
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d = 2.62 m
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Find:
vi = ??
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vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)
0 m2/s2 = vi2 - 51.35 m2/s2
51.35 m2/s2 = vi2
vi = 7.17 m/s
Return to Problem 10
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Given:
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a = -9.8 m/s2
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vf = 0 m/s
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d = 1.29 m
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Find:
vi = ??
t = ??
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vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)
0 m2/s2 = vi2 - 25.28 m2/s2
25.28 m2/s2 = vi2
vi = 5.03 m/s
To find hang time, find the time to the peak and then double it.
vf = vi + a*t
0 m/s = 5.03 m/s + (-9.8 m/s2)*tup
-5.03 m/s = (-9.8 m/s2)*tup
(-5.03 m/s)/(-9.8 m/s2) = tup
tup = 0.513 s
hang time = 1.03 s
Return to Problem 11
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Given:
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vi = 0 m/s
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vf = 521 m/s
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d = 0.840 m
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Find:
a = ??
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vf2 = vi2 + 2*a*d
(521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)
271441 m2/s2 = (0 m/s)2 + (1.68 m)*a
(271441 m2/s2)/(1.68 m) = a
a = 1.62*105 m /s2
Return to Problem 12
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Given:
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a = -9.8 m/s2
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vf = 0 m/s
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t = 3.13 s
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Find:
d = ??
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-
(NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)
First use: vf = vi + a*t
0 m/s = vi + (-9.8 m/s2)*(3.13 s)
0 m/s = vi - 30.7 m/s
vi = 30.7 m/s (30.674 m/s)
Now use: vf2 = vi2 + 2*a*d
(0 m/s)2 = (30.7 m/s)2 + 2*(-9.8 m/s2)*(d)
0 m2/s2 = (940 m2/s2) + (-19.6 m/s2)*d
-940 m2/s2 = (-19.6 m/s2)*d
(-940 m2/s2)/(-19.6 m/s2) = d
d = 48.0 m
Return to Problem 13
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Given:
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vi = 0 m/s
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d = -370 m
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a = -9.8 m/s2
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Find:
t = ??
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d = vi*t + 0.5*a*t2
-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2
-370 m = 0+ (-4.9 m/s2)*(t)2
(-370 m)/(-4.9 m/s2) = t2
75.5 s2 = t2
t = 8.69 s
Return to Problem 14
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Given:
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vi = 367 m/s
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vf = 0 m/s
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d = 0.0621 m
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Find:
a = ??
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vf2 = vi2 + 2*a*d
(0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)
0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a
-134689 m2/s2 = (0.1242 m)*a
(-134689 m2/s2)/(0.1242 m) = a
a = -1.08*106 m /s2
(The - sign indicates that the bullet slowed down.)
Return to Problem 15
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Given:
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a = -9.8 m/s2
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t = 3.41 s
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vi = 0 m/s
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Find:
d = ??
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d = vi*t + 0.5*a*t2
d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2
d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)
d = -57.0 m
(NOTE: the - sign indicates direction)
Return to Problem 16
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Given:
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a = -3.90 m/s2
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vf = 0 m/s
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d = 290 m
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Find:
vi = ??
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vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)
0 m2/s2 = vi2 - 2262 m2/s2
2262 m2/s2 = vi2
vi = 47.6 m /s
Return to Problem 17
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Given:
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vi = 0 m/s
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vf = 88.3 m/s
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d = 1365 m
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Find:
a = ??
t = ??
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vf2 = vi2 + 2*a*d
(88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)
7797 m2/s2 = (0 m2/s2) + (2730 m)*a
7797 m2/s2 = (2730 m)*a
(7797 m2/s2)/(2730 m) = a
a = 2.86 m/s2
vf = vi + a*t
88.3 m/s = 0 m/s + (2.86 m/s2)*t
(88.3 m/s)/(2.86 m/s2) = t
t = 30. 8 s
Return to Problem 18
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Given:
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vi = 0 m/s
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vf = 112 m/s
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d = 398 m
|
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Find:
a = ??
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vf2 = vi2 + 2*a*d
(112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)
12544 m2/s2 = 0 m2/s2 + (796 m)*a
12544 m2/s2 = (796 m)*a
(12544 m2/s2)/(796 m) = a
a = 15.8 m/s2
Return to Problem 19
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Given:
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a = -9.8 m/s2
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vf = 0 m/s
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d = 91.5 m
|
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Find:
vi = ??
t = ??
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First, find speed in units of m/s:
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m)
0 m2/s2 = vi2 - 1793 m2/s2
1793 m2/s2 = vi2
vi = 42.3 m/s
Now convert from m/s to mi/hr:
vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)
vi = 94.4 mi/hr
Return to Problem 20