Physics 2514
Lecture 5
P. Gutierrez
Department of Physics & Astronomy
University of Oklahoma
Physics 2514 – p. 1/16
Goal
The goal for today’s lecture is to derive the kinematic
expressions for constant acceleration in 1-dimension.
Physics 2514 – p. 2/16
Brief Review
In the last lecture we generalized the concept of average velocity
to instantaneous velocity (velocity at an instant of time).
Take the limit as the time interval goes to zero
dx(t)
∆x
=
∆t→0 ∆t
dt
v(t) = lim
PSfrag replacements
x0
v(t) represents the slope tangent to a
function (plot) of position vs time at the
point x(t).
The velocity is the differential change in
position dx at time t in a time differential
time interval dt
x
∆x
t
∆t
t0
Physics 2514 – p. 3/16
Position from Velocity
Given a relation between the position and time, we can
determine the velocity by calculating the derivative. Let’s now
see how to go in the opposite direction, given a relation between
velocity and time, how do we calculate the position.
Start with relation for
average velocity
vavg
vx
Area = v0 ∆t = ∆x
v0
∆x
replacements
⇒ ∆x
= PSfrag
= v0 ∆t
∆t
For constant velocity
∆x is area under the
curve
∆t
t
Physics 2514 – p. 4/16
Position from Velocity
v(t)
Let’s generalize to the case of non-constant velocity
Take general function v(t)
and approximate as
150
series of constant velocity
100
segments
PSfrag replacements 50
Displacement is sum of
0
segments
0 1 2 3 4
Pn
t
∆x = i=1 vi ∆t
5
6
Take limit lim∆t→0
Area = lim
∆t→0
n
X
i=1
vi ∆t =
Z
tf
v(t)dt = ∆x = xf − xi
ti
Physics 2514 – p. 5/16
Constant Velocity
Let’s consider the case of constant velocity
Assume that the velocity is vx0
Carry out integral for general expression use indefinite
integral
Z t
x(t) =
vx0 dt = vx0 [t − t0 ] + x(t0 )
t0
In most cases we take t0 = 0 then x(t) = vx0 t + x0
Yields equation of a straight line with slope (initial velocity)
vx0 and intercept (initial position) x0
This equation holds for any straight line motion, ∴
generalize to s(t) = vs0 ∆t + s(t0 )
Physics 2514 – p. 6/16
Clicker
The position of an object is given by x = 2t2 − 7t. What is the
instantaneous velocity of the object when t = 4 s. The units of x
are meters.
1. 4 m/s
2. 8.3 m/s
3. 2 m/s
4. 9 m/s
5. 6.5 m/s
Physics 2514 – p. 7/16
Instantaneous Acceleration
As in the case of velocity, we start with the average acceleration
to start our discussion of instantaneous acceleration.
The average acceleration to instantaneous acceleration
aavg =
∆v
∆t
⇒
dv(t)
∆x
=
∆t→0 ∆t
dt
a(t) = lim
PSfrag replacements
v0
The average acceleration represents a slope on a velocity versus
time plot.
a(t) is the differential change in
velocity dv at time t in a differential time interval dt
v
∆v
t
∆t
t0
Physics 2514 – p. 8/16
Instantaneous Acceleration
As a starting point in the study of acceleration, and the most
common case for this course, we will consider constant
acceleration.
This refers to the case when the velocity increases by the same
amount over successive time intervals.
Note this case is a special case, and does not hold for motion in
general.
Physics 2514 – p. 9/16
Position vs Time
150
x
Motion Diagram
PSfrag replacements
1
100
50
y
0
0
0
1
2
3
4
5
6
t
Velocity vs Time
-1
0
50
100
x
∆t same for all intervals.
150
v
placements
Constant Acceleration
PSfrag replacements
60
50
40
30
20
10
1
2
3
4
5
6
t
Physics 2514 – p. 10/16
Constant Acceleration
Based on the previous slide, constant acceleration corresponds
to a linear relation between velocity and time. The relation is:
v(t) = a[t − t0 ] + v(t0 ) = at + v0
the 2nd equation is the case when we take t0 = 0
Where a represents the slope, and v0 the intercept (the value of
the velocity at t = 0)
Recall that the acceleration is given by the derivative of the
velocity with respect to time:
d
dv
= (at + v0 ) = a
dt
dt
as expected
Physics 2514 – p. 11/16
Constant Acceleration
Let’s now derive an expression between position and time for
the case of constant acceleration.
Begin with the expression for average acceleration, which is
also exactly correct for constant acceleration
∆v
a=
∆t
⇒
v(t) = at + v0
v
To calculate the position, find the
PSfrag replacements
area under the curve
Area = x − x0 =
a∆t
1
a[∆t]2 + v0 ∆t
2
where we use ∆t = t − t0 where
t0 = 0.
v0
∆t
t
Physics 2514 – p. 12/16
Constant Acceleration
The expression on the previous slide can also be arrived at
using calculus
Start with the expression for velocity under the assumption
of constant acceleration
v(t) =
dx
= at + v0
dt
Next integrate
Z
tf
ti
dx
dt =
dt
Z
tf
ti
1
(at + v0 ) dt ⇒ x−x(t0 ) = a[∆t]2 +v0 ∆t
2
1 2
⇒ x(t) = at + v0 t + x0
2
Physics 2514 – p. 13/16
Constant Acceleration
At times it is useful to have a relation between the position and
velocity without the time.
Start with the equations for velocity and time and solve for
the time
v − v0
v = at + v0 ⇒ t =
a
Substitute t into the expression for the position

v − v0

t=

a
1 2

x = at + v0 t + x0 
2
⇒
v 2 − v02 = 2a(x − x0 )
Physics 2514 – p. 14/16
Summary
The following are the kinematic relations derived for systems under
constant acceleration with straight line motion.
Velocity and time
v = a∆t + v0
Position and time
1
s = a∆t2 + v0 ∆t + s0
2
Position and velocity
v 2 = v02 + 2a(s − s0 )
With v0 the velocity at t = t0 , and s0 the position at t = t0 .
Physics 2514 – p. 15/16
Assignment
Read sections 2.6 and 2.7 for the next lecture
Tuesday’s lecture will cover examples of systems having
constant acceleration.
Physics 2514 – p. 16/16