1. Linear Acceleration
Objectives:
• Introduce the concept of linear acceleration
• Learn how to compute and estimate linear
acceleration
• Understand the difference between average
and instantaneous acceleration
• Learn to use the laws of constant acceleration
Change in Velocity
• Because velocity is a vector, a change in velocity
is also a vector
• Change in velocity has magnitude and direction
vy (m/s)
∆v = change
from vinitial to vfinal
vintial
vfinal
vx (m/s)
1

2. Computing a Change in Velocity
• Compute change in velocity (∆v) by vector subtraction
∆v = vfinal – vinitial
vy (m/s)
vinitial
vfinal
vx (m/s)
–vinitial –vinitial
∆v
Linear Acceleration
• The rate of change of linear velocity
• Acceleration is a vector; has magnitude and direction
change in velocity
acceleration =
change in time
• Shorthand notation:
v2 – v1 ∆v
a = =
t2 – t1 ∆t
• Has units of length/time2 (e.g. m/s2, ft/s2)
2

3. Computing Acceleration
• direction of accel. = direction of change in velocity
• magnitude of accel. = magnitude of change in velocity
change in time
• component of accel. = component of change in velocity
change in time
y ∆v y a = ∆v / ∆t
/ ∆t
∆v
∆v
∆vy ay = ∆vy / ∆t
=
θ a θ
x x
∆vx ax = ∆vx / ∆t
Acceleration as Change in Speed
• Acceleration can reflect a change in the
magnitude of velocity without a change in
direction (e.g. a change in speed)
vy ay
v2 a = ∆v / ∆t
∆v
v1
θ θ
vx ax
3

4. Acceleration as Change in Direction
• Acceleration can reflect a change in the direction
of velocity without a change in magnitude
vy ay
v1
∆v
v v2 ax
v
vx
a = ∆v / ∆t
Acceleration in General
• In general, acceleration reflects a change in both the
magnitude and direction of velocity
• Can picture as accel. along original direction of motion
and accel. away from original direction of motion
vy ay a along dir. of motion
a away from
∆v dir. of motion
v1
v2
a = ∆v / ∆t
vx ax
4

5. Example Problem #1
A basketball player is moving forward at 6 m/s at
the start of a jump.
He leaves the ground 250 ms later with a forward
velocity of 2.5 m/s and an upward velocity of
4 m/s
What was his average acceleration between
starting the jump and leaving the ground?
Acceleration in 1-D
• Velocity and acceleration in same direction:
magnitude of velocity increases
• Velocity and acceleration in opposite direction:
magnitude of velocity decreases (deceleration)
Velocity Acceleration Change in Velocity
(+) (+) Increase in + dir.
(+) (–) Decrease in + dir.
(–) (–) Increase in – dir.
(–) (+) Decrease in – dir.
5

6. Example Problem #2
The instantaneous vertical velocity was determined at
intervals of 100 ms during the movement shown.
What was the average vertical acceleration over each
100 ms interval?
1.4 Time Velocity
1.2 (s) (m/s)
vertical position (m)
1 0 0
0.8 0.1 4
0.6
0.2 6
0.4
0.3 3
0.2
0.4 -1
0
0 0.1 0.2 0.3 0.4 0.5 0.5 -5
time (s)
Acceleration as a Slope
• Graph x-component of velocity vs. time
• x-component of acceleration from t1 to t2
= slope of the line from vx at t1 to vx at t2
vx (m/s)
Slope : ∆vx / ∆t = a x
∆ vx
∆t
t1 t2 time (s)
6

7. Average vs. Instantaneous Accel.
• Previous formulas give us the average acceleration
between an initial time (t1) and a final time (t2)
• Instantaneous acceleration is the acceleration at a
single instant in time
• Can estimate instantaneous acceleration using the
central difference method:
v (at t1 + ∆t) – v (at t1 – ∆t)
a (at t1) =
2 ∆t
where ∆t is a very small change in time
Instantaneous Accel. as a Slope
• Graph of x-component of velocity vs. time
slope = instantaneous
x-acceleration at t1
vx (m/s)
slope = average
x-acceleration
from t1 to t2
∆t
t1 t2 time (s)
7

8. Estimating Acceleration from Velocity
vx (m/s) Identify points with
zero slope = points
with zero acceleration
0 Portions of the curve
time (s) with positive slope
have positive accel.
(i.e. acceleration in
ax (m/s2)
the + direction)
Portions of the curve
with negative slope
0 have negative accel.
time (s)
(i.e. acceleration in
the – direction)
Example Problem #3
Below are velocity vs. time plots for two sprinters in
the 100 m. Graph the acceleration of each sprinter.
Who had the greater peak velocity? Acceleration?
14
12
10
Velocity (m/s)
8
6
4
JOHNSON
2
LEWIS
0
0 2 4 6 8 10
Time (s)
8

9. Laws of Constant Acceleration
• Under conditions where acceleration is constant:
v2 = v 1 + a * ∆t
d = v 1 * ∆ t + (½) a * ( ∆ t)2
v22 = v 12 + 2 a * d
where:
a = acceleration
v 1 = velocity at initial (or first) time t1
v 2 = velocity at final (or second) time t2
d = displacement (change in position) between t1 and t2
∆ t = change in time (= t2 – t1 )
Use + values for + direction, – values for – direction
Special Cases
• If acceleration (a) = 0:
v 2 = v1 + a * ∆ t v 2 = v1
d = v1 * ∆t + (½) a * ( ∆t)2 d = v1 * ∆ t
• If initial velocity (v1) = 0:
v 2 = v1 + a * ∆ t v2 = a * ∆t
d = v1 * ∆t + (½) a * ( ∆t)2 d = (½) a * ( ∆t)2
v22 = v12 + 2 a * d v22 = 2 a * d
• If final velocity (v2) = 0:
v 2 = v1 + a * ∆ t v1 = – a * ∆t
v22 = v12 + 2 a * d v12 = – 2 a * d
9

10. Example Problem #4
A cyclist starts down a hill at 2 m/s and accelerates
down the road at a constant rate of 0.6 m/s 2
If the road down the hill is 200 m long, how fast is
she moving at the bottom?
If she then uses her brakes to decelerate at a
constant 1.2 m/s 2, how much time will it take her
to return to a velocity of 2 m/s?
How far will she travel while she is decelerating?
10