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Lesson 1: Describing Motion with Words

Introduction to the Language of Kinematics

Scalars and Vectors

Distance and Displacement

Speed and Velocity

Acceleration


Lesson 2: Describing Motion with Diagrams

Introduction to Diagrams

Ticker Tape Diagrams

Vector Diagrams


Lesson 3: Describing Motion with Position vs. Time Graphs

The Meaning of Shape for a p-t Graph

The Meaning of Slope for a p-t Graph

Determining the Slope on a p-t Graph


 

Lesson 4: Describing Motion with Velocity vs. Time Graphs

The Meaning of Shape for a v-t Graph

The Meaning of Slope for a v-t Graph

Relating the Shape to the Motion

Determining the Slope on a v-t Graph

Determining the Area on a v-t Graph

 

Lesson 5: Free Fall and the Acceleration of Gravity

Introduction to Free Fall

The Acceleration of Gravity

Representing Free Fall by Graphs

How Fast? and How Far?

The Big Misconception

 

Lesson 6: Kinematic Equations

The Kinematic Equations

Problem-Solving

Kinematic Equations and Free Fall

Sample Problems and Solutions

Kinematic Equations and Graphs

 

 

Lesson 6: Kinematic Equations and Problem-Solving

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 which 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

  1. 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.

    See solution below.

     

  2. 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.

    See solution below.

     

  3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

    See solution below.

     

  4. 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.

    See solution below.

     

  5. 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.

    See solution below.

     

  6. 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 1.8 seconds, then what is the acceleration and what is the distance which the sled travels?

    See solution below.

     

  7. 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.

    See solution below.

     

  8. An engineer is designing the runway for an airport. Of the planes which 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?

    See solution below.

     

  9. 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).

    See solution below.

     

  10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

    See solution below.

     

  11. 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)?

    See solution below.

     

  12. 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).
    See solution below.

     

  13. 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.)
    See solution below.

     

  14. 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.
    See solution below.

     

  15. 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.)
    See solution below.

     

  16. 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.
    See solution below.

     

  17. It was once recorded that a Jaguar left skid marks which 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.
    See solution below.

     

  18. 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.
    See solution below.

     

  19. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.
    See solution below.

     

  20. 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.
See solution below.



 

 

 

Solutions to Above Problems

  1. Given:

    a = +3.2 m/s2

    t = 32.8 s

    vi = 0 m/s

    Find:

    d = ??
    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

     

  2. Given:

    d = 110 m

    t = 5.21 s

    vi = 0 m/s

    Find:

    a = ??
    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

     

  3. Given:

    a = -9.8 m

    t = 2.6 s

    vi = 0 m/s

    Find:

    d = ??

    vf = ??

    d = vi*t + 0.5*a*t2

    d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

    d = 33 m

    vf = vi + a*t

    vf = 0 + (-9.8 m/s2)*(2.6 s)

    vf = -25.5 m/s (- indicates direction)

    Return to Problem 3

     

  4. Given:

    vi = 18.5 m/s

    vf = 46.1 m/s

    t = 2.47 s

    Find:

    d = ??

    a = ??

    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

     

  5. Given:

    vi = 0 m/s

    d = -1.40 m

    a = -1.67 m/s2

    Find:

    t = ??
    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

     

  6. Given:

    vi = 0 m/s

    vf = 44 m/s

    t = 1.80 s

    Find:

    a = ??

    d = ??

    a = (Delta v)/t

    a = (444 m/s - 0 m/s)/(1.80 s)

    a = 247 m/s2

    d = vi*t + 0.5*a*t2

    d = (0 m/s)*(1.80 s)+ 0.5*(247 m/s2)*(1.80 s)2

    d = 0 m + 400 m

    d = 400 m

    (Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

    Return to Problem 6

     

  7. Given:

    vi = 0 m/s

    vf = 7.10 m/s

    d = 35.4 m

    Find:

    a = ??
    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

     

  8. Given:

    vi = 0 m/s

    vf = 65 m/s

    a = 3 m/s2

    Find:

    d = ??
    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

     

  9. Given:

    vi = 22.4 m/s

    vf = 0 m/s

    t = 2.55 s

    Find:

    d = ??
    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

     

  10. Given:

    a = -9.8 m/s2

    vf = 0 m/s

    d = 2.62 m

    Find:

    vi = ??
    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

     

  11. Given:

    a = -9.8 m/s2

    vf = 0 m/s

    d = 1.29 m

    Find:

    vi = ??

    t = ??

    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

     

  12. Given:

    vi = 0 m/s

    vf = 521 m/s

    d = 0.840 m

    Find:

    a = ??
    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

     

  13. Given:

    a = -9.8 m/s2

    vf = 0 m/s

    t = 3.13 s

    Find:

    d = ??
    1. (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time.)
    First use: vf = vi + a*t

    0 m/s = vi + (-9.8 m/s2)*(3.13 s)

    0 m/s = vi - 30.6 m/s

    vi = 30.6 m/s

    Now use: vf2 = vi2 + 2*a*d

    (0 m/s)2 = (30.6 m/s)2 + 2*(-9.8 m/s2)*(d)

    0 m2/s2 = (938 m/s) + (-19.6 m/s2)*d

    -938 m/s = (-19.6 m/s2)*d

    (-938 m/s)/(-19.6 m/s2) = d

    d = 47.9 m

    Return to Problem 13

     

  14. Given:

    vi = 0 m/s

    d = -370 m

    a = -9.8 m/s2

    Find:

    t = ??
    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

     

  15. Given:

    vi = 367 m/s

    vf = 0 m/s

    d = 0.0621 m

    Find:

    a = ??
    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

     

  16. Given:

    a = -9.8 m/s2

    t = 3.41 s

    vi = 0 m/s

    Find:

    d = ??
    d = vi*t + 0.5*a*t2

    d = (0 m/s)*(2.6 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

     

  17. Given:

    a = -3.90 m/s2

    vf = 0 m/s

    d = 290 m

    Find:

    vi = ??
    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

     

  18. Given:

    vi = 0 m/s

    vf = 88.3 m/s

    d = 1365 m

    Find:

    a = ??

    t = ??

    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

     

  19. Given:

    vi = 0 m/s

    vf = 112 m/s

    d = 398 m

    Find:

    a = ??
    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

     

  20. Given:

    a = -9.8 m/s2

    vf = 0 m/s

    d = 91.5 m

    Find:

    vi = ??

    t = ??

    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

 

 

Lesson 6: Kinematic Equations and Problem-Solving

 

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