Chapter 2

Motion Along a Straight Line

2.1 Displacement, Time, and Average Velocity

1D motion. Very often it is convenient to model an object whose motion
you analyze (e.g. car, runner, stone, etc.) as a point particle. Then to
describe motion of the object we can use a vector in some coordinate system.
If the motion is along a straight line, then it is convenient (to choose x-axis of
the coordinate system) to lie along the direction of motion. The origin may
be chosen at the location of the object in the initial position, or anywhere
along the line of motion.
Position. As time progresses the position vector (or location of the
object) will change with time. Thus, position vector is nothing but three
functions of time
⃗r(t) = (x(t), y(t), z(t)). (2.1)
If motion is only along x-axis, then the position is only a single function of
time
⃗r(t) = (x(t), 0, 0). (2.2)
In this chapter we are only interested in motion along a straight line and
thus the position will be denoted by x(t). For example,

x(t) = 7 m,
! "
t
x(t) = sin 2 m + 5 m,
1s
# m$
x(t) = t · 5
# sm $ # m$
2
x(t) = t · 3 2 + t · 4 +5m (2.3)
s s

15
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 16

Average velocity. Whatever the function x(t) is one can always find
displacement
∆x ≡ x(t2 ) − x(t1 ) (2.4)
during time interval
∆t ≡ t2 − t1
and average velocity as

∆x x(t2 ) − x(t1 )
vavg ≡ = . (2.5)
∆t t2 − t1
So if you know the positions of your object x(t1 ) and x(t2 ) at two different
moments of time t1 and t2 , then you can always calculate average velocity.
For example, if a runner finished a 10 km race in 1 hour, then his/her average
velocity was
x(1h) − x(0h) 10km km
vavg = = = 10 . (2.6)
1h − 0h 1h h
Since ∆t is always positive, the displacement ∆x and thus average velocity
can be ⎧
⎨> 0 motion mostly in the direction of + î
⎪
vavg < 0 motion mostly in the direction of − î (2.7)
⎪
= 0 for any round-trip motion.
⎩

Question: Can traveled distance be larger than displacement? Can it be

smaller? Can it be the same?

2.2 Instantaneous Velocity

Instantaneous velocity. Average velocity is telling you how on average the
position of your object is changing during time interval ∆t. Instantaneous
velocity is the same thing, but when the time interval is taken to zero, i.e.
∆x dx
v(t) ≡ lim vavg = lim = . (2.8)
∆t→0 ∆t→0 ∆t dt
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 17

Thus given x as a function of t one can always find instantaneous velocity at

every moment of time by simple differentiation. For example,
dx m
x(t) = 7 m, → v(t) = =0
! " dt s! "
t dx t m
x(t) = sin 2 m + 5 m → v(t) = = cos 2
1s dt 1s s
# m$ dx m
x(t) = t · 5 → v(t) = =5
s dt s
# m$ # m$ dx # m$ m
x(t) = t2 · 3 2 + t · 4 + 5 m → v(t) = = t · 6 2 + 4 (2.9)
.
s s dt s s
Graphical representation of both types of velocities should be clear: average
velocity is a slope of a line joining the coordinates of initial (x1 , t1 ) and final
(x2 , t2 ) points points on x(t) graph; instantaneous velocity if a slope of the
tangent line to x(t) at a given time.

2.3 Average and Instantaneous acceleration

Average acceleration. Given instantaneous velocity as a function of time
v(t), one can calculate average acceleration.

∆v v(t2 ) − v(t1 )
aavg ≡ = . (2.10)
∆t t2 − t1
This is analogous to how the average velocity was defined from (instan-
taneous) position. For example if a car was going with velocity v(t1 ) =
100 km/hand then one hour later its velocity is 150 km/h, then average
acceleration must have been
150 km/h − 100 km/h km
aavg = = 50 2 . (2.11)
1h h
This is completely independent on what the instantaneous velocity was dur-
ing this hour.
Instantaneous acceleration. By drawing an analogy with instanta-
neous velocity, one might guess that instantaneous acceleration is defined by
taking a limit,
∆v dv
a(t) ≡ lim aavg = lim = . (2.12)
∆t→0 ∆t→0 ∆t dt
Thus instantaneous acceleration is an acceleration at a given moment of time.
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 18

For example,
m dv m
v(t) = 0 , → a(t) = =0
! " s dt s2 ! "
t m dv t m
v(t) = cos 2 → a(t) = = − sin 2 2
1s s dt 1s s
m dv m
v(t) = 5 → a(t) = =0 2
s dt s
# m$ m dv m
v(t) = t · 6 2 + 4 → a(t) = =6 2 (2.13)
s s dt s
Graphical representation of both types of accelerations should be clear: av-
erage velocity is a slope of a line joining the coordinates of initial (v(t1 ), t1 )
and final (v(t2 ), t2 ) points on v(t) graph; instantaneous velocity if a slope of
the tangent line to v(t) at a given time.

2.4 Motion with Constant Acceleration

Constant acceleration. Motion with constant acceleration is a motion for
which instantaneous acceleration is a constant function

a(t) = ax , (2.14)

where the subscript x is to remind you that the motion is along x axis only.
Velocity for such motion changes as

vx (t) = v0x + ax t (2.15)

and position changes as

1
x(t) = x0 + v0x t + ax t2 . (2.16)
2
Note that ax , v0x , x0 are some fixed numbers representing acceleration, ve-
locity and position at time t = 0, but t is a variable which can take any
non-negative value. the fact that Eqs. (2.14), (2.15) and (2.16) are self-
consistent can be checked by differentiation,

dx(t)
v(t) = = v0x + ax t
dt
dv(t)
a(t) = = ax . (2.17)
dt
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 19

Useful relations. One can obtain other useful relations from Eqs. (2.15)
and (2.16) by expressing t in one equation and plugging it into another. For
example, Eq. (2.15) implies
vx (t) − v0x
t= (2.18)
ax
and by plugging it in Eq. (2.16) we get
! " ! "2
vx (t) − v0x 1 vx (t) − v0x
x(t) = x0 + v0x + ax
ax 2 ax
2
vx (t) · v0x − v0x vx (t)2 − 2vx (t)v0x + v0x2
x(t) − x0 = +
ax 2ax
2 2 2
2ax (x(t) − x0 ) = 2vx (t) · v0x − 2v0x + vx (t) − 2vx (t)v0x + v0x
2ax (x(t) − x0 ) = vx (t)2 − v0x 2
(2.19)

or
vx (t)2 = v0x
2
+ 2ax (x(t) − x0 ) . (2.20)
Another useful relation is obtained from Eqs. (2.15) and (2.16) by expressing
ax in one equation and plugging it into another. For example, Eq. (2.15)
implies
vx (t) − v0x
ax = (2.21)
t
and by plugging it in Eq. (2.16) we get
! "
1 vx (t) − v0x 2
x(t) = x0 + v0x t + t
2 t
1 1
x(t) = x0 + v0x t + vx (t)t − v0x t
2 2
1 1
x(t) − x0 = vx (t)t + v0x t (2.22)
2 2
or
1
x(t) − x0 = (vx (t) + v0x ) t. (2.23)
2
Question: What would change in equations (2.14), (2.15), (2.16), (2.20)
and (2.23) if the initial time is at t0 ̸= 0?
Example 2.4. A motorcyclist heading east through a small town accel-
erates at a constant acceleration 4.0 m/s2 after he leaves city limits. At time
t = 0 he is 5.0 m east of the city-limits while he moves east at 15 m/s. (a)
Find position and velocity at t = 2.0 s. (b) Where is he when his speed is
25 m/s?
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 20

Step 1: Coordinate system: Where should the x-axis point? Pointing

east with origin at the city-limit.
Step 2: Initial conditions: What is known about initial conditions?

t0 = 0s
x0 = 5.0 m
v0x = 15 m/s
ax = 4.0 m/s2 . (2.24)

Step 3: Final conditions: (a) What is position and velocity at t = 2.0 s?

Using Eq. (2.16) we get

1
x(2.0 s) = (5.0 m) + (15 m/s) (2.0 s) + 4.0 m/s2 (2.0 s)2 = 43 m (2.25)
2
and using Eq. (2.15)

vx (2.0 s) = 15 m/s + 4.0 m/s2 (2.0 s) = 23 m/s.

) *
(2.26)

(b) Where is he when his speed is 25 m/s? Using Eq. (2.20) we get

vx (t)2 − v0x
2
x(t) = x0 + (2.27)
2ax
or by substituting the known quantities

(25 m/s) 2 − (15 m/s) 2

x(t) = (5.0 m) + = 55 m. (2.28)
2 4.0 m/s2
) *

2.5 Free Falling Bodies

Free falling. Free fall is a very deep concept in physics. All it means
that in 4-dimensional space-time free falling object move along straight lines
(geodesics). For now we are only interested in motion very close to surface of
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 21

the Earth and such motions can be approximated as motions with constant
acceleration
ax = g = 9.80 m/s2 . (2.29)
Thus all of the concepts and equations considered in the previous section
apply. Note that g is taken to be positive and thus it makes sense to choose
x-axis to point vertically and downward. If you choose the x-axis to point
upward then ax = −9.80 m/s2 .
Example 2.6. One-euro coin is dropped from the Leaning Tower of Pisa
and falls free from rest. What are its position and velocity after 1.0 s?

Step 1. Coordinate system: Where should the x-axis point? Let’s assume
that x-axis points vertically and downwards (this is opposite to what the book
assumes) with origin at the top of Pisa tower.
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 22

Step 2. Initial condition: What is known about initial conditions?

t0 = 0s
x0 = 0m
v0x = 0 m/s
ax = +9.80 m/s2 . (2.30)

Step 3. Final conditions: What are its position and velocity after 1.0 s?
Using Eq. (2.16) we get

1
x(1.0 s) = (0 m) + (0 m/s) (1.0 s) + 9.80 m/s2 (1.0 s)2 = 4.9 m. (2.31)
2
and using Eq. (2.15)

vx (1.0 s) = 0 m/s + +9.80 m/s2 (1.0 s) = 9.80 m/s.

) *
(2.32)

Example 2.7. You throw a ball vertically upward from the roof of a tall
building. The ball leaves your hand at a point even with the roof railing with
an upward speed of 15.0 m/s; the ball is then in free fall. On its way back,
it just misses the railing. Find (a) the ball’s position and velocity 1.00 s and
4.00 s after leaving your hand; (b) the ball’s velocity when it is 5.00 m above
the railing; (c) the maximum height reached; (d) the ball’s acceleration when
it is at its maximum height.
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 23

Step 1: Coordinate system: Choose x-axis to point vertically and upward

with origin at the roof railing.
Step 2: Initial conditions: What is known about initial conditions?

t0 = 0s
x0 = 0m
v0x = 15.0 m/s
ax = −9.80 m/s2 . (2.33)

Step 3: Final conditions: (a) What is ball’s position and velocity t = 1.00 s
and t = 4.00 s after leaving your hand? Using Eq. (2.16) we get

1)
−9.80 m/s2 (1.0 s)2 = 10.1 m
*
x(1.00 s) = (0 m) + (15.0 m/s) (1.0 s) +
2
1)
−9.80 m/s2 (4.0 s)2 = −18.4
*
x(4.00 s) = (0 m) + (15.0 m/s) (4.0 s) + (2.34)
m
2
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 24

and using Eq. (2.15)

vx (1.0 s) = 15.0 m/s + −9.80 m/s2 (1.0 s) = 5.2 m/s
) *

vx (4.0 s) = 15.0 m/s + −9.80 m/s2 (4.0 s) = −24.2 m/s. (2.35)

) *

(b) What is the ball’s velocity when it is 5.00 m above the railing? Using
Eq. (2.20)
vx (t)2 = v0x
2
+ 2ax (x(t) − x0 ) . (2.36)
we get +
vx = (15.0 m/s)2 + 2 (−9.80 m/s2 ) (5.00 m − 0 m) (2.37)
with two solutions
vx = +11.3 m/s (2.38)
or
vx = −11.3 m/s. (2.39)
(c) What is the maximum height reached? Using Eq. (2.27)

(0 m/s)2 − (15.0 m/s)2

x = (0 m) + = 11.5 m. (2.40)
2 −9.80 m/s2
) *

(d) What is the ball’s acceleration when it is at its maximum height? We

are considering motion with constant acceleration and thus the acceleration
is always
ax = −9.80 m/s2 (2.41)
where the minus sign (for our choice of coordinate system, see Step 1) indi-
cates that the acceleration points downwards.

2.6 Velocity and Position by Integration.

Velocity by integration. An expression for velocity can always be obtain
given acceleration as a function of time. From definition of acceleration
dvx (t)
ax (t) = (2.42)
dt
and thus in differential from
dvx = ax (t)dt (2.43)
or in integral form
, vx (t2 ) , t2
dvx = ax (t)dt. (2.44)
vx (t1 ) t1
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 25

Therefore for t1 = 0 and t2 = T we have

, T
vx (T ) = vx (0) + ax (t)dt. (2.45)
0

If we replace

T → t
t → τ
vx (0) → v0x (2.46)

then we get , t
vx (t) = v0x + ax (τ )dτ (2.47)
0
and in the case of constant acceleration

vx (t) = v0x + ax t. (2.48)

Position by integration. Similarly, an expression for position can be

obtain given velocity as a function of time. From definition of velocity
dx(t)
vx (t) = (2.49)
dt
and thus in differential from

dx = vx (t)dt (2.50)

or in integral form
, x(t2 ) , t2
dx = vx (t)dt. (2.51)
x(t1 ) t1

Therefore for t1 = 0 and t2 = T we have

, T
x(T ) = x(0) + vx (t)dt. (2.52)
0

If we replace

T → t
t → τ
x(0) → x0 (2.53)
CHAPTER 2. MOTION ALONG A STRAIGHT LINE 26

then we get , t
x(t) = x0 + vx (τ )dτ (2.54)
0
and in the case of constant acceleration
, t
x(t) = x0 + (v0x + ax τ )dτ
0
1
x(t) = x0 + v0x t + ax t2 . (2.55)
2
Example 2.9. Sally is driving along a straight highway in her 1965
Mustang. At t = 0, when she is moving at 10 m/s in the positive x-direction,
she passes a signpost at x = 50 m. Her x-acceleration as a function of time
is
ax (t) = 2.0 m/s2 − 0.10 m/s3 t.
) *
(2.56)
(a) Find her x-velocity vx (t) and x(t) as functions of time. (b) When is her
x-velocity greatest? (c) What is that maximum x-velocity? (d) Where is the
car when it reaches that maximum x-velocity?

Step 1: Coordinate system: The coordinate system was already chosen for
us with x-axis pointing along the direction of motion and origin at x = 0 m
signpost.
Step 2: Initial conditions: What is known about initial conditions?

t0 = 0
x0 = 50 m
v0x = 10 m/s
a0x = 2.0 m/s2 . (2.57)

Step 3: Final conditions: (a) What is her x-velocity and position as a

CHAPTER 2. MOTION ALONG A STRAIGHT LINE 27

function of time? By substituting Eq. (2.56) into Eq. (2.44) we get

, t
2.0 m/s2 − 0.10 m/s3 τ dτ
) ) * *
vx (t) = (10 m/s) +
0
1
vx (t) = 10 m/s + t · 2.0 m/s2 − t2 · 0.10 m/s3
2
vx (t) = 10 m/s + t · 2.0 m/s2 − t2 · 0.05 m/s3 (2.58)

and from Eq. (2.54) we get

, t
10 m/s + t · 2.0 m/s2 − t2 · 0.05 m/s3 dτ
) *
x(t) = (50 m) +
0
1 1
x(t) = 50 m + t · 10 m/s + t2 · 2.0 m/s2 − t3 · 0.05 m/s3
2 3
x(t) ≈ 50 m + t · 10 m/s + t2 · 1.0 m/s2 − t3 · 0.017 m/s3 (2.59)

(b) When is her x-velocity greatest? The maximum of a function occurs

if the slope changes from positive to negative. This happens in an instance
when the slope is exactly zero, i.e.

dvx (tmax )
= ax (tmax ) = 2.0 m/s2 − tmax · 0.10 m/s3 = 0 (2.60)
dt
or when
2.0 m/s2
tmax = = 20 s. (2.61)
0.10 m/s3
(c) What is that maximum x-velocity? Using Eq. (2.58) we get

vx (20 s) = 10 m/s + 20 s · 2.0 m/s2 − (20 s)2 · 0.05 m/s3 = 30 m/s. (2.62)

(d) Where is the car when it reaches that maximum x-velocity? Using
Eq. (2.59) we get

x(20 s) = 50 m+(20 s)·10 m/s+(20 s)2 ·1.0 m/s2 −(20 s)3 ·0.017 m/s3 = 517 m.
(2.63)