ACCELERATION AND
FREE FALL
CHAPTER 3
ACCELERATION
• The changing of an object’s velocity over time.
• Average Acceleration
• 𝑎=
∆𝑣
∆𝑡
=
𝑣𝑓 −𝑣𝑖
𝑡𝑓 −𝑡𝑖
• SI Unit: m/s2
• Vector Quantity
ACCELERATION
• When velocity and acceleration are in the same
direction, the speed of the object increases with
time.
• When velocity and acceleration are in opposite
directions, the speed of the object decreases with
time.
ACCELERATION
• 𝑎=
∆𝑣
∆𝑡
=
𝑣𝑓 −𝑣𝑖
𝑡𝑓 −𝑡𝑖
=
−10.0𝑚 𝑠− −2.0𝑚 𝑠
4.0𝑠
= −2.0 𝑚 𝑠 2
• The camel is not slowing down, it’s velocity is
increasing in the negative-x direction.
4.0s
vf=-10.0m/s
vi=-2.0m/s
ACCELERATION
• 𝑎=
∆𝑣
∆𝑡
=
𝑣𝑓 −𝑣𝑖
𝑡𝑓 −𝑡𝑖
=
−2.0𝑚 𝑠− −10.0𝑚 𝑠
4.0𝑠
= 2.0 𝑚 𝑠 2
• The camel is slowing down, it’s velocity and
acceleration are in opposite directions.
4.0s
vf=-2.0m/s
vi=-10.0m/s
DECELERATION VS.
NEGATIVE ACCELERATION
• Deceleration – Reduction in speed
• Negative Acceleration – Acceleration vector is in
the negative-x direction
INSTANTANEOUS ACCELERATION
• The limit of the average acceleration as the time
interval ∆t approaches zero.
∆𝑣
∆𝑡→0 ∆𝑡
• 𝑎 = lim
MOTION DIAGRAMS
ONE-DIMENSIONAL MOTION WITH
CONSTANT ACCELERATION
• Many applications of mechanics involve objects
moving with constant acceleration
• Constant Acceleration
• Instantaneous acceleration = average acceleration
COMPARING MOTION GRAPHS
VELOCITY AS A FUNCTION OF TIME
• 𝑎=
𝑣𝑓 −𝑣𝑖
𝑡𝑓 −𝑡𝑖
• where:
𝑡𝑖 = 0
𝑣 = 𝑣0 , 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑡 = 0
𝑣𝑓 = 𝑣, 𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑎𝑛𝑦 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑡𝑖𝑚𝑒 𝑡
• 𝑎=
𝑣−𝑣0
𝑡
• 𝑣 = 𝑣0 + 𝑎𝑡
REALITY CHECK
• A sports car moving at constant speed travels 110m in 5.0 s. If
it then brakes and comes to a stop in 4.0 s, what is its
acceleration in Express the answer in terms of “g’s,” where
1.00𝑔 = 9.80𝑚/𝑠 2
• The initial velocity of the car is the average speed of the car
before it accelerates.
• 𝑣=
∆𝑥
∆𝑡
=
110𝑚
5.0𝑠
= 22 𝑚 𝑠 = 𝑣0
• The final speed is 0 and the time to stop is 4.0 s.
• 𝑎=
𝑣−𝑣0
𝑡
=
0−22𝑚 𝑠
4.0𝑠
= −5.5 𝑚 𝑠 2
1𝑔
9.80𝑚 𝑠 2
= −0.56𝑔
DISPLACEMENT AS A
FUNCTION OF TIME
1
2
• ∆𝑥 = 𝑣0 𝑡 + 𝑎𝑡 2
The area under the line on a velocity vs. time
graph is equal to the displacement of the
object.
AN AUTOMOBILE MANUFACTURER CLAIMS THAT ITS SUPERDELUXE SPORTS CAR WILL ACCELERATE UNIFORMLY FROM
REST TO A SPEED OF 38.9M/S IN 8.00S.
• a. Determine the acceleration of the car.
• 𝑎=
𝑣−𝑣0
𝑡
=
38.9𝑚 𝑠
8.00𝑠
= 4.86 𝑚 𝑠 2
• b. Find the displacement of the car in the first
8.00s.
• ∆𝑥 =
1
𝑣0 𝑡 + 𝑎𝑡 2
2
=0
1
+
2
4.86 𝑚 𝑠 2 8.00𝑠
2
= 156𝑚
VELOCITY AS A FUNCTION OF
DISPLACEMENT
•
𝑣2
= 𝑣0
2
+ 2𝑎∆𝑥
EQUATIONS FOR MOTION IN A STRAIGHT
LINE UNDER CONSTANT ACCELERATION
Equation
Information Given by Equation
𝑣 = 𝑣0 + 𝑎𝑡
Velocity as a function of time
1 2
∆𝑥 = 𝑣0 𝑡 + 𝑎𝑡
2
Displacement as a function of time
2
𝑣 = 𝑣0
2
+ 2𝑎∆𝑥
Velocity as a function of displacement
IN COMING TO A STOP, A CAR LEAVES SKID MARKS 92.0M LONG
ON THE HIGHWAY. ASSUMING A DECELERATION OF 7.00M/S2,
ESTIMATE THE SPEED OF THE CAR JUST BEFORE BRAKING.
•
•
•
•
𝑣=0
∆𝑥 = 92.0𝑚
𝑎 = −7.00 𝑚 𝑠 2
𝑣0 =?
2
• 𝑣 = 𝑣0
2
+ 2𝑎∆𝑥
• 𝑣0 = 𝑣 2 − 2𝑎∆𝑥
• 𝑣0 =
0 − 2 −7.00 𝑚 𝑠 2 92.0𝑚 = 35.9𝑚/𝑠
FREE FALL
• When air resistance is ignored, all objects in free fall
near the Earth’s surface fall at the same constant
acceleration.
• Galileo
• Dropped objects of different weights of
Leaning Tower of Pisa
• Or did he?
• Inclined Planes
• Diluting Gravity
FREE FALL
• Any object moving under the
influence of gravity alone
• Does not need to start from rest
• An object thrown up into the air is in
free fall even when its altitude is
increasing.
• Free Fall Acceleration (g)
•
•
•
•
Varies slightly with latitude
Decreases as altitude increases
9.80 𝑚 𝑠 2
“Up” =positive 𝑦 direction
• 𝑎 = −𝑔 = −9.80 𝑚 𝑠 2
FREE FALL
• A tennis player on serve tosses a ball straight up.
While the ball is in free fall , does the acceleration
•
•
•
•
•
A. Increase
B. Decrease
C. Increase then Decrease
D. Decrease then Increase
E. Remain Constant
FREE FALL
• A tennis player on serve tosses a ball straight up.
While the ball is in free fall , does the acceleration
•
•
•
•
•
A. Increase
B. Decrease
C. Increase then Decrease
D. Decrease then Increase
E. Remain Constant
FREE FALL
• As the tennis ball in the previous question travels
through the air, its speed
•
•
•
•
•
A. Increases
B. Decreases
C. Decreases then Increases
D. Increases then Decreases
E. Remains the Same
FREE FALL
• As the tennis ball in the previous question travels
through the air, its speed
•
•
•
•
•
A. Increases
B. Decreases
C. Decreases then Increases
D. Increases then Decreases
E. Remains the Same
A BASEBALL IS HIT NEARLY STRAIGHT UP
INTO THE AIR WITH A SPEED OF 22M/S.
• a. How high does it go?
• 𝑣 2 = 𝑣0
2
+ 2𝑎 𝑦 − 𝑦0 , 𝑦 = 𝑦0 +
𝑣 2 −𝑣0
2𝑎
2
• where, 𝑦0 = 0, 𝑣0 = 22 𝑚 𝑠 , 𝑣 = 0, 𝑎 = −9.80 𝑚 𝑠 2
• 𝑦 =0+
0− 22𝑚 𝑠 2
2 −9.80𝑚 𝑠 2
= 24.7𝑚
• b. How long is it in the air?
1
2
1
2
• ∆𝑦 = 𝑣0 𝑡 + 𝑎𝑡 2 = 0, 𝑡 𝑣0 + 𝑎𝑡 = 0, 𝑡 = 0, 𝑡 =
• 𝑡=
−2 22𝑚 𝑠
−9.8𝑚 𝑠 2
= 4.50𝑠
−2𝑣0
𝑎