Motion with Constant Acceleration
1 Particle Under Constant Acceleration
In the case of motion with constant acceleration, the instantaneous and average accelerations are equal so we have
ax = a x =
vxf − vxi
∆vx
=
.
∆t
tf − t i
If we let ti = 0 and tf = t, we get
ax =
vxf − vxi
.
t
Solving for vxf we get an expression for the final velocity in terms of the initial velocity,
the acceleration, and the time
vxf = vxi + ax t.
(1)
The average velocity can be written as
vx =
1
(vxi + vxf )
2
or
vx =
xf − x i
.
t
Solving the latter expression for xf we get
xf = xi + vx t.
Substituting the first expression for vx into the above equation yields an expression for the
final position in terms of the initial position, the initial velocity, the final velocity, and the
time
1
xf = xi + (vxi + vxf ) t
(2)
2
Substituting eqn. (1) into eqn. (2), we get
xf = x i +
1
(vxi + vxi + ax t) t
2
which yields an expression for the final position in terms of the initial position, the initial
velocity, the acceleration, and the time
1
xf = xi + vxi t + ax t2 .
2
1
(3)
Solving eqn. (1) for t and substituting into (2), we get
xf = x i +
1
vxf − vxi
(vxi + vxf )
2
a
which can be solved for the final velocity squared in terms of the initial velocity squared,
the acceleration, and the displacement
(4)
2
2
vxf
= vxi
+ 2ax (xf − xi ) .
Graphs of position, velocity, and acceleration as a function of time for motion with constant acceleration are shown in Figure 1.
1.1 Constant Acceleration Example
You brake your Porsche from 75 km/h to 45 km/h over a distance of 88 m. (a) What is the
acceleration, assumed to be constant?
vxi = 75 km/h vxf = 45 km/h xf − xi = 88 m ax =?
2
2
vxf
= vxi
+ 2ax (xf − xi )
2
2
vxf
− vxi
(45 km/h)2 − (75 km/h)2
=
= −2.0 × 104 km/h2
ax =
2 (xx − xi )
2 (0.088 km)
ax = −2.0 ×
km
104 2
h
1000 m
1 km
1h
3600 s
!2
= −1.5 m/s2
(b) What is the elapsed time?
xf − x i =
t=
1
(vxi + vxf ) t
2
2 (xf − xi )
2 (0.088 km)
=
= 1.5 × 10−3 h = 5.3 s
vxi + vxf
75 km/h + 45 km/h
(c) If you continue to slow down with the acceleration calculated in (a), how much time
will elapse in bringing the car to rest from 75 km/h?
vxi = 75 km/h
vxf = 0 ax = −2.0 × 104 km/h2
t =?
vxf = vxi + ax t
t=
0 − 75 km/h
vxf − vxi
=
= 3.8 × 10−3 h = 14 s
ax
−2.0 × 104 km/h2
2
Figure 1: Graphs of position, velocity, and acceleration versus time for motion with constant acceleration.
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2 Freely Falling Objects
Free fall near the surface of the earth is the most common example of motion with (nearly)
constant acceleration. Free-fall acceleration near the surface of the earth is g = 9.80 m/s 2 .
In solving free-fall problems, we will use the equations derived earlier for motion with
constant acceleration with the following substitutions
x → y and ax → ay = −g = 9.8 m/s2 .
2.1 Free Fall Example
A worker drops a wrench down an elevator shaft of a tall building. (a) Where is the wrench
1.5 seconds later?
yi = 0
vyi = 0
t = 1.5 s yf =?
1
yf = yi + vyi t + ay t2
2
1
−9.8 m/s2 (1.5 s)2 = −11 m
yf = 0 + 0 +
2
(b) How fast is the wrench falling after 1.5 s?
vyf = vyi + ay t = 0 + −9.8 m/s2 (1.5 s) = −15 m/s
3 Homework Set 2 - Due Monday, Sept. 13
• Read Sections 2.6-2.7
• Answer Question 9 & 11
• Do Problems 19, 26, 33 & 39
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