Physics 10:11- Sample Problems and Solutions

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 runway at 3.20 m/s2 for 32.8 s until is !nally lifts off the ground. Determine the distance traveled before takeoff. See Answer

See solution below.(http://www.physicscl below.(http://www.physicsclassroom.com/Class/1DKi assroom.com/Class/1DKin/U1L6d.cfm#sol1) n/U1L6d.cfm#sol1)

2. A car starts from rest and accelerates uniformly uniformly over a time of 5.21 seconds for a distance distance of 110 m. Determine the acceleration of the car. See Answer


3. Upton Chuck is riding the Giant Drop at Great America. America. If Upton free falls for 2.60 seconds, what will be his !nal velocity and how far will he fall? See Answer


4. A race car accelerates uniformly from from 18.5 m/s to 46.1 m/s m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. See Answer


5. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of 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 Answer

See solution below.

6. Rocket-powered sleds are used to test the human human response to acceleration. acceleration. If a rocket-powered sled is accelerated to a speed speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels? See Answer


7. A bike accelerates uniformly uniformly from rest to a speed of 7.10 m/s m/s over a distance of 35.4 m. Determine the acceleration acceleration of the bike. See Answer


8. 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? See Answer

See solution below.(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol8)

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 Answer


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


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 Answer


12. A bullet leaves a ri"e with a muzzle velocity of 521 m/s. While accelerating through the barrel of the ri "e, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration). See Answer


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 Answer


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 Answer

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 Answer


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 Answer


17. 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. See Answer


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 Answer


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 Answer


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 !eld)? Assume negligible air resistance. See Answer


Solutions to Above Problems 1. Given: a = +3.2 m/s2

Find: t = 32.8 s

vi = 0 m/s 2

d = vi*t + 0.5*a*t d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2 )*(32.8 s)2 d = 1720 m Return to Problem 1

d = ??

Given: d = 110 m

Find: t = 5.21 s

vi = 0 m/s

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. Find:

Given: a = -9.8 m

t = 2.6 s

d = ?? vf = ??

vi = 0 m/s

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

4. Find:

Given: vi = 18.5 m/s

vf = 46.1 m/s

d = ?? a = ??

t = 2.47 s

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 v f2 = v i2 + 2*a*d) Return to Problem 4(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q4)

5. Given: vi = 0 m/s

Find: d = -1.40 m

a = -1.67 m/s2 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

t = ??

6. Find:

Given: vi = 0 m/s

vf = 444 m/s

a = ?? d = ??

t = 1.83 s 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 v f2 = v i2 + 2*a*d) Return to Problem 6(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q6)


Find: vf = 7.10 m/s

d = 35.4 m

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q7)


Find: a = 3 m/s2

vf = 65 m/s

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q8)

9. Given: vi = 22.4 m/s

Find: vf = 0 m/s

t = 2.55 s

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q9)

d = ??

Given:

Find:

a = -9.8 m/s2

vf = 0 m/s

d = 2.62 m

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q10)

11. Find:

Given: a = -9.8 m/s2

vf = 0 m/s

vi = ?? t = ??

d = 1.29 m

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 !nd hang time, !nd 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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q11)

12. Given:

Find:

vi = 0 m/s

vf = 521 m/s

d = 0.840 m

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q12)

13. Given:

Find: 2

a = -9.8 m/s a.

vf = 0 m/s

t = 3.13 s

d = ??

(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)

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q13)

14. Given:

Find:

vi = 0 m/s

a = -9.8 m/s2

d = -370 m

t = ??

2

d = vi*t + 0.5*a*t -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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q14)

15. Given:

Find:

vi = 367 m/s

vf = 0 m/s

d = 0.0621 m

a = ??

2

vf = vi2 + 2*a*d 2

(0 m/s)2 = (367 m/s) + 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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q15)

16. Given:

Find: 2

a = -9.8 m/s

t = 3.41 s

vi = 0 m/s

d = ??

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q16)

17. Given:

Find:

a = -3.90 m/s2

vf = 0 m/s

d = 290 m

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q17)

18. Find:

Given: vi = 0 m/s

vf = 88.3 m/s

a = ?? t = ??

d = 1365 m 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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q18)


Find: vf = 112 m/s

d = 398 m

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q19)

20. Find:

Given: a = -9.8 m/s2

vf = 0 m/s

d = 91.5 m First, !nd 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 = ?? t = ??

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(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q20)

Next Section: Kinematic Equations and Graphs(/class/1DKin/Lesson-6/Kinematic-Equations-and-Graphs) 302

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Physics 10:11- Sample Problems and Solutions

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