Kinematic Practice Questions and Solutions

Kinematics Practice Questions and Answers

1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before take-off.

2. A car starts from rest and accelerates uniformly for 5.21 seconds over a distance of 1.10 x 102 m. Determine the acceleration of the car.

3. 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.

4. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled accelerates to a speed of 444 m/s in 1.8 seconds, what is its acceleration and how far does it travel?

5. 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.

6. An engineer is designing the runway of an airport. Of the planes which will use the airport, the lowest acceleration rate is 3 m/s2 and the lowest take-off speed is 65 m/s. Assuming these minimum parameters are for the same plane, what is the minimum allowed length of the runway?

7. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car.

8. A bullet leaves a rifle with a muzzle velocity of 52 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.

9. A bullet is moving with a speed of 67 m/s when it enters a hill of dirt. The bullet penetrates a distance of 6.21 m. Determine the acceleration of the bullet while moving into dirt.

10. A car left skid marks which were 2.90 x 102 m in length. Assuming that the car skidded to a stop with a constant acceleration of –3.90 m/s2, determine the speed of the car before it began to skid.

1 11. A plane has a take-off speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time taken to reach the take-off speed.

12. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration of the dragster.

13. A river's current is moving from west to east at a speed of 5.0 m/s. A boat wishes to head north and to not be blown off course. The boat is trying to maintain 10.0 m/s north, but needs to travel into the current to maintain a heading due north. With what speed is the boat actually travelling as it heads due north, and at what angle relative to the shore must the boat point?

14. An Air Canada plane is travelling at 1.000 x 103 km/h in a direction 40o east of north. a) Find the components of the velocity vector by finding the component in the northerly direction and the easterly direction. b) How far to the north and how far to the east would the plane travel in 4 hours?

15. A captain walks 4.0 km/h directly across a cruise ship whose speed relative to the earth is 12.0 km/h. What is the speed of the captain with respect to the earth?

2 Solutions

1. a = 3.20 m/s2 t = 32.8 s

vi = 0 m/s d = ? *3 Sig. Digs. 2 ∆d = vi∆t + ½ a∆t d = (0 m/s)(32.8 s)+ 0.5(3.20 m/s2)(32.8 s)2 d = 1721 m or 1.72 x 103 m

2. d = 1.10 x 102 m t = 5.21 s

vi = 0 m/s a = ? *3 Sig. Digs. 2 ∆d = vi∆t + ½ a∆t 1.10 x 102 m = (0 m/s)(5.21 s)+ 0.5(a)(5.21 s)2 1.10 x 102 m = (13.57 s2)a a = (1.10 x 102 m)/(13.57 s2) a = 8.10 m/s2

3. vi = 18.5 m/s

vf = 46.1 m/s t = 2.47 s a = ? d = ? *3 Sig. Digs.

vf = vi + a∆t

a = vf - vi ∆t

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

3 a = 11.2 m/s2 or 1.12 x 101 m/s2

2 ∆d = vi∆t + ½ a∆t 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 or 7.98 x 101 m

2 2 (Note: d can also be calculated using the following equation: vf = vi + 2ad)

4. vi = 0 m/s

vf = 44 m/s t = 1.80 s a = ? d = ? *2 Sig. Digs.

vf = vi + a∆t

a = vf - vi ∆t

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

a = 246.666 m/s2 or 2.5 x 102 m/s2

2 d = vit + ½at

d = (0 m/s)(1.80 s)+ 0.5(246.666 m/s2)(1.80 s)2

d = 0 m + 399.589 m

4 d = 399.589 m d = 4.0 x 102 m

5. vi = 0 m/s

vf = 7.10 m/s d = 35.4 m a = ? *3 Sig. Digs. 2 2 vf = vi + 2ad

(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 or 7.12 x 10-1 m/s2

6. vi = 0 m/s

vf = 65 m/s a = 3 m/s2 d = ? *1 Sig. Dig. 2 2 vf = vi + 2ad

(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

5 d = 704 m or 7 x 102 m

7. vi = 22.4 m/s

vf = 0 m/s t = 2.55 s d = ? *3 Sig. Digs. (There are several ways to do this question)

vav = ∆d (Remember average velocity = (vi + vf) ∆t 2

∆d = vav∆t

d = (vi + vf)t 2 d = (22.4 m/s + 0 m/s)(2.55 s) 2 d = (11.2 m/s)(2.55 s)

d = 28.6 m or 2.86 x 101 m

8. vi = 0 m/s

vf = 52 m/s d = 0.840 m a = ? *2 Sig. Digs. 2 2 vf = vi + 2ad

(52 m/s)2 = (0 m/s)2 + 2(a)(0.840 m)

2704 m2/s2 = (0 m/s)2 + (1.68 m)a

6 (2704 m2/s2)/(1.68 m) = a

a = 1609 m/s2 or 1.6 x103 m/s2 (This may seem an absurd acceleration, but it makes sense if you calculate the time it took to travel this short distance... ie. about 0.03 seconds :)

9. vi = 67 m/s

vf = 0 m/s d = 6.21 m a = ? *2 Sig. Digs. 2 2 vf = vi + 2ad

(0 m/s)2 = (67 m/s)2 + 2(a)(6.21 m)

0 m2/s2 = (4489 m2/s2) + (12.42 m)a

-4489 m2/s2 = (12.42 m)a

(-4489 m2/s2)/(12.42 m) = a

a = -361.43 m/s2

a = -3.6 x 102 m/s2

10. a = –3.90 m/s2

vf = 0 m/s d = 2.90 x 102 m

vi = ?

7 *3 Sig. Digs. 2 2 vf = vi + 2ad

2 2 2 2 (0 m/s) = vi + 2(-3.90 m/s )( 2.90 x 10 m)

2 2 2 2 2 0 m /s = vi - 2262 m /s

2 2 2 2262 m /s = vi

2 2 2 vi = (2262 m /s )

1 vi = 47.6 m/s or 4.76 x 10 m/s

11. vi = 0 m/s

vf = 88.3 m/s d = 1365 m a = ? t = ? *3 Sig. Digs. 2 2 vf = vi + 2ad

(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

8 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 or 3.08 x 101 s

12. vi = 0 m/s

vf = 112 m/s d = 398 m a = ? *3 Sig. Digs. 2 2 vf = vi + 2ad

(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 or 1.58 x 101 m/s2

9 *2 Sig. Figs. 2 2 2 vBS + vWS = vBW 2 2 2 vBS = vBW - vWS 2 2 2 vBS = (10.0) - (5.0) 2 vBS = 100 - 25

vBS = 8.66 m/s or 8.7 m/s

14. *4 Sig. Figs. a) The component in the northerly direction is: (1.000 x 103 km/h)(cos 40o) = 766.044 km/h or 7.660 x 102 km/h The component in the easterly direction is: (1.000 x 103 km/h)(sin 40o) = 642.787 km/h or 6.428 x 102 km/h

b) Distance travelled... use d = vt The distance travelled in the northerly direction:

10 d = (766.044)(4) = 3064 km or 3.064 x 103 km The distance in the easterly direction: d = (642.787)(4) = 2571 km or 2.571 x 103 km

15. *2 Sig. Figs.

vCS = velocity of captain with respect to ship

vSE = velocity of ship with respect to earth

vCE = velocity of captain with respect to earth 2 2 2 vCE = vCS + vSE 2 2 2 vCE = (4.0) + (12.0) 2 vCE = 16 + 144 2 vCE = 160 1 vCE = 12.649 km/h or 1.3 x 10 km/h