grade 9 module

1. 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.
See Answer
See solution below.
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 Answer
See solution below.
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?
See Answer
See solution below.
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 Answer

2. See solution below.
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 Answer
See solution below.
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?
See Answer
See solution below.
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 Answer
See solution below.

3. 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.
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
See solution below.
A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the
kangaroo.
See Answer
See solution below.

4. 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
See solution below.
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 Answer
See solution below.
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
See solution below.
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.

5. See Answer
See solution 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.)
See Answer
See solution below.
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
See solution below.
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

6. See solution below.
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
See solution below.
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
See solution below.
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 Answer
See solution below.

7. Solutions to Above Problems
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
Given:

8. 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
Given:

9. 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)*(2.60 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

10. 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

11. d = 79.8 m
(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
Return to Problem 4
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

12. 1.68 s2 = t2
t = 1.29 s
Return to Problem 5
Given:
vi = 0 m/s
vf = 444 m/s
t = 1.83 s
Find:
a = ??
d = ??
a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)

13. 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
Given:
vi = 0 m/s
vf = 7.10 m/s
d = 35.4 m
Find:

14. 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
Given:
vi = 0 m/s
vf = 65 m/s
a = 3 m/s2
Find:

15. 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
Given:
vi = 22.4 m/s
vf = 0 m/s
t = 2.55 s
Find:
d = ??

16. 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
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)

17. 0 m2/s2 = vi2 - 51.35 m2/s2
51.35 m2/s2 = vi2
vi = 7.17 m/s
Return to Problem 10
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)

18. 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

19. 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
Given:

20. a = -9.8 m/s2
vf = 0 m/s
t = 3.13 s
Find:
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)
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

21. -940 m2/s2 = (-19.6 m/s2)*d
(-940 m2/s2)/(-19.6 m/s2) = d
d = 48.0 m
Return to Problem 13
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

22. -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
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)

23. 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
Given:
a = -9.8 m/s2
t = 3.41 s
vi = 0 m/s
Find:

24. 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
Given:
a = -3.90 m/s2
vf = 0 m/s
d = 290 m
Find:
vi = ??

25. 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
Given:
vi = 0 m/s
vf = 88.3 m/s
d = 1365 m
Find:
a = ??
t = ??

26. 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
Given:

27. 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

28. 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:

29. vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)
vi = 94.4 mi/hr
Return to Problem 20