C H A P T E R 3

ACCELERATION AND FREE FALL

ACCELERATION

• The changing of an object’s velocity over time.

• Average Acceleration

• ത𝑎 =∆𝑣

∆𝑡=

𝑣𝑓−𝑣0

𝑡𝑓−𝑡0

• 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

• ത𝑎 =∆𝑣

∆𝑡=

𝑣𝑓−𝑣0

𝑡𝑓−𝑡0=

−10.0 Τ𝑚 𝑠− −2 Τ.0𝑚 𝑠

4.0𝑠= −2.0 Τ𝑚 𝑠2

• The camel is not slowing down, it’s velocity is

increasing in the negative-x direction.

4.0s

vf=-10.0m/s v0=-2.0m/s

ACCELERATION

• ത𝑎 =∆𝑣

∆𝑡=

𝑣𝑓−𝑣0

𝑡𝑓−𝑡0=

−2.0 Τ𝑚 𝑠− −10 Τ.0𝑚 𝑠

4.0𝑠= 2.0 Τ𝑚 𝑠2

• The camel is slowing down, it’s velocity and

acceleration are in opposite directions.

4.0s

v0=-10.0m/svf=-2.0m/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.

• 𝑎 = lim∆𝑡→0

∆𝑣

∆𝑡

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−𝑣0

𝑡𝑓−𝑡0

• where:

𝑡0 = 0𝑣 = 𝑣0, 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑡 = 0𝑣𝑓 = 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/𝑠2

• The initial velocity of the car is the average speed of the car before it accelerates.

• 𝑣 =∆𝑥

∆𝑡=

110𝑚

5.0𝑠= 22 Τ𝑚 𝑠 = 𝑣0

• The final speed is 0 and the time to stop is 4.0 s.

• 𝑎 =𝑣−𝑣0

𝑡=

0−22 Τ𝑚 𝑠

4.0𝑠= −5.5 Τ𝑚 𝑠2

1𝑔

9.80 Τ𝑚 𝑠2= −0.56𝑔

DISPLACEMENT AS AFUNCTION OF TIME

• ∆𝑥 = 𝑣0𝑡 +1

2𝑎𝑡2

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 SUPER-DELUXE 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=

38.9 Τ𝑚 𝑠

8.00𝑠= 4.86 Τ𝑚 𝑠2

• b. Find the displacement of the car in the first

8.00s.

• ∆𝑥 = 𝑣0𝑡 +1

2𝑎𝑡2 = 0 +

1

24.86 Τ𝑚 𝑠2 8.00𝑠 2 = 156𝑚

VELOCITY AS A FUNCTION OF DISPLACEMENT

• 𝑣2 = 𝑣02 + 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

∆𝑥 = 𝑣0𝑡 +1

2𝑎𝑡2 Displacement as a function of time

𝑣2 = 𝑣02 + 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 Τ𝑚 𝑠2

• 𝑣0 =?

• 𝑣2 = 𝑣02 + 2𝑎∆𝑥

• 𝑣0 = 𝑣2 − 2𝑎∆𝑥

• 𝑣0 = 0 − 2 −7.00 Τ𝑚 𝑠2 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 Τ𝑚 𝑠2

• “Up” =positive 𝑦 direction

• 𝑎 = −𝑔 = −9.80 Τ𝑚 𝑠2

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?

• 𝑣2 = 𝑣02+ 2𝑎𝑦, 𝑦 =

𝑣2−𝑣02

2𝑎

• where, 𝑦0 = 0, 𝑣0 = 22 Τ𝑚 𝑠 , 𝑣 = 0, 𝑎 = −9.80 Τ𝑚 𝑠2

• 𝑦 = 0 +0− 22 Τ𝑚 𝑠 2

2 −9.80 Τ𝑚 𝑠2= 24.7𝑚

• b. How long is it in the air?

• ∆𝑦 = 𝑣0𝑡 +1

2𝑎𝑡2 = 0, 𝑡 𝑣0 +

1

2𝑎𝑡 = 0, 𝑡 = 0, 𝑡 =

−2𝑣0

𝑎

• 𝑡 =−2 22 Τ𝑚 𝑠

−9.8 Τ𝑚 𝑠2= 4.50𝑠