Unit 92 ms 2 average acceleration a D v D t 50 ms 25ms 4s 2s 125 ms 2 B C A A C none B n f n i t f t i Practice A pool ball travelling at 25 ms towards the cushion bounces off at 15 ms If the ball was in contact with the ID: 778692
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Slide1
Calculating Acceleration
Unit 9.2
Slide2m/s
2
average acceleration
Slide3a =
D
v
D
t
50 m/s – 25m/s
4s – 2s
=
=
12.5 m/s
2
Slide4B
C
A
A
C
none
B
Slide5n
f
–
n
i
t
f
–
ti
Slide6Practice
: A pool ball travelling at 2.5 m/s towards the cushion bounces off at 1.5 m/s. If the ball was in contact with the cushion for 0.20 s, what is the ball’s acceleration? (Assume towards the cushion is the positive direction.)
= - 20 m/s
2
Slide7Practice
: A car accelerates from rest at 3.0 m/s
2
forward for 5.0 s. What is the velocity of the car at the end of 5.0 s?
a =
D
v
v
f
- v
i
D
t
D
t
a
D
t
+ v
i
= (3.0m/s
2
)(5.0s) + 0m/s
=
v
f
=
= 15 m/s [forward]
v
i
= 0 m/s
D
t
= 5.0s
a = 3.0 m/s
2
v
f
= ?
Slide8Practice
: A train is travelling west at 14 m/s. How long would to increase its velocity to 22 m/s west, if it accelerated at 0.50 m/s
2
west? Assign east direction positive (+).
a =
D
v
v
f
- v
i
D
t
D
t
v
f
- v
i
= -22m/s – (-14m/s)
=
D
t
=
= 16 s
v
i
= -14 m/s
v
f
= -22 m/s
a = -0.50 m/s
2
D
t
= ?
a
-0.50m/s
2
Slide9Practice
: Solve the following problems.
A truck starting from rest accelerates uniformly to 18 m/s [W] in 4.5 s. What is the truck’s acceleration?
a =
D
v
v
f
- v
i
D
t
Dt
= -18m/s – 0 m/s
=
= - 4 m/s
2
v
i
= 0 m/s
v
f
= -18 m/s
D
t
= 4.5 s
a = ?
4.5 s
Slide10Practice
: Solve the following problems.
A toboggan moving 5.0 m/s forward decelerates backward at -0.40 m/s
2
for 10 s. What is the toboggan’s velocity at the end of the 10 s?
a =
D
v
v
f
- v
i
D
t
Dt
a
D
t
+ v
i
= (-0.40m/s
2
)(10s) + 5m/s
=
v
f
=
= 1 m/s
v
i
= 5 m/s
a = -0.40 m/s
2
D
t
= 10 s
v
f
= ?
Slide11Practice
: Solve the following problems.
How much time does it take a car travelling south at 12 m/s to increase its velocity to 26 m/s south if it accelerates at 3.5 m/s
2
south?
a =
D
v
v
f
- v
i
D
t
Dt
v
f
- v
i
= -26m/s – (-12m/s)
=
D
t
=
= 4 s
D
t
= ?
v
i
= -12 m/s
v
f
= -26 m/s
a = -3.5 m/s
2
a
-3.5m/s
2
Slide12Air resistance
same rate
9.8 m/s
2
-9.8 m/s
2
Slide13Slide14A → B
B
B → C
Velocity
Acceleration
decreasing
-9.8 m/s
2
0
-9.8 m/s
2
increasing
-9.8 m/s
2
Slide15Practice
: Solve the following problems.
Suppose a rock falls from the top of a cliff. What is the change in velocity of the rock after it has fallen for 1.5 s? Assign “down” as negative (-).
What is the change in velocity of a brick that falls for 3.5 s?
Slide16Practice
: Solve the following problems.
A ball is thrown straight up into the air at 14 m/s. How long does it take for the ball to slow down to an upward velocity of 6.0 m/s?
A rock is thrown downwards with an initial velocity of 8.0 m/s. What is the velocity of the rock after 1.5 s?