Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 2
Motion Along a Straight Line
Younes Sina

2. The motion of a particle along a straight
line at a constant acceleration is called
the "Uniformly Accelerated Motion."
a= constant Uniformly Accelerated Motion

3. Velocity is defined as the change in displacement per unit of time.
t=t0
t=t
Δx

5. V=0 km/h
V=260 km/h
A value of 9 (km/h)/s means the velocity changes by 9.0 km/h
during each second of the motion
Example
/
t= 29 s

7. V=0 km/s V= 260 km/s
V=0 m/s V= 72 m/s
An acceleration of 2.5 m/s2 is read as “2.5 meters per second per second” (or “2.5
meters per second squared”) and means that the velocity changes by 2.5 m/s
during each second of the motion.

8. Negative Acceleration and Decreasing Velocity
A drag racer crosses the finish line, and the driver deploys a
parachute and applies the brakes to slow down (above Figure).
The driver begins slowing down when t0 =9.0 s and the car’s
velocity is v0= 28 m/s. When t= 12.0 s, the velocity has been
reduced to v= 13 m/s. What is the average acceleration of the
dragster?
v


10. Example:
A car travels a distance of 350 miles in
5.0 hours. Find its average speed.

11. V=Δx/Δt=350 mi/5.0hr=70.0 mi/hr
Example : A car's speed changes from 15m/s to 25m/s in
along a straight road while moving in one way.
Find the magnitude of the acceleration of the car
during 5 seconds.

12. A=(vf-vi/tf-ti)=(25m/s-15 m/s)/5.0 s=2.0 m/s2
if
if
tt
vv
a



2/0.2
0.5
/15/25
sm
s
smsm
a 



13. The Equation of Motion of Uniformly Accelerated Motion:
x=(1/2) at2+vit
where x and t are variables, and a and vi, the constant

14. Example : A car traveling along a straight road at 8.0m/s
accelerates to a speed of 15.0m/s in 5.0s. Determine
(a) its acceleration,
(b) the distance it travels during this period,
(c) its equation of motion,
(d) the interval of validity of this equation, and
(e) the distance already traveled at t = 2.0 s.
v= 8.0 m/s v= 15.0 m/s
x

15. x = (1/2) a t2 + vi t
Calculation of acceleration using final and initial
velocities and time:
a = (vf -vi ) /t
Calculation of traveled distance using
acceleration, initial velocity and time:
Any object that moves along a straight line and at constant
acceleration, has an equation of motion in the form:

16. Part (e) x(2.0) = 0.7 (2.0)2 + 8.0 (2.0) = 18.8m ≈ 19m.
Part (a):
a = (15.0 m/s - 8.0 m/s) / 5.0s = 1.4m/s2.
Part (b):
x = (1/2) (1.4m/s2) (5.0s)2 + (8.0m/s)(5.0s) = 58m (rounded to 2 sig. figs.)
The equation of motion of this car can be found by substituting the constants in the equation.
The constants are: vi = 8.0m/s and a = 1.4m/s2.
The equation of motion becomes:
Part (c) x(t) = (1/2)(1.4)t2 + 8.0t or x(t) = 0.7t2 + 8.0t
Part (d) This equation is valid for 0 ≤ t ≤ 5.0s only.
(The statement of the problem gives us information for a 5.0s period only).

17. Example: A cyclist traveling at 30.0m/s along a straight road
comes uniformly to stop in 5.00s. Determine the stopping
acceleration, the stopping distance, and the equation of
motion.
v= 30.0 m/s v= 0 m/s
t=5.00 s

18. Solution:
a = (vf - vi ) / t
a= ( 0 - 30.0m/s) / 5.00s = -6.00 m/s2.
x = (1/2) a t2 + vi t
x = (1/2)(-6.00m/s2)(5.00s)2 + (30.0m/s)(5.00s)
x= +75m.
x = (1/2) (-6.00m/s2)t2 + (30.0m/s)t
x(t) = -3.00t2 + 30.0t.
(a relation between x and t ).

19. Example : The equation of motion of a lady skiing along a steep
and straight ramp is x = 3.40t2 + 2.10t where x is in meters
and t in seconds. Determine
(a) her distance from the starting point at the end of
1.0s, 2.0s, 3.0s, 4.0s, and 5.0s. periods.
(b) Determine the distances she travels during the 1st, 2nd, 3rd,
4th, and 5th seconds.
(c) Are the answers in Part (b) equal? If not why?
What should be the pace of motion for such distances to be equal?

20. 1.0 s
2.0 s
3.0 s
t= 0
x = 3.40t2 + 2.10t

21. x(0) = 3.40(0)2 + 2.10(0) = 0.
x(1) = 3.40(1)2 + 2.10(1) = 5.50m.
x(2) = 3.40(2)2 + 2.10(2) = 17.8m.
x(3) = 3.40(3)2 + 2.10(3) = 36.9m.
x(4) = 3.40(4)2 + 2.10(4) = 62.8m.
x(5) = 3.4(5)2 + 2.1(5) = 95.5m.
d1 = x(1) - x(0) = 05.5m - 00.0m = 5.5m.
d2 = x(2) - x(1) = 17.8m - 05.5m = 12.3m.
d3 = x(3) - x(2) = 36.9m - 17.8m = 19.1m.
d4 = x(4) - x(3) = 62.8m - 36.9m = 25.9m.
d5 = x(5) - x(4) = 95.5m - 62.8m = 32.7m.
d1 is the distance traveled in the 1st second.
d2 is the distance traveled in the 2nd second,
and so on.

22. (c) As can be seen from Part (b), the distances traveled in
subsequent seconds become greater and greater.
What is this telling us?
It is because the motion is an accelerated one.
In accelerated motion, unequal distances are traveled in
equal time intervals.

23. x = (1/2) a t2 + vi t

24. An Equation With No Apparent Time Element
a = (vf -vi) / t → t=(vf -vi) /a
x = (1/2)at2 + vit
a
vv
v
a
vv
ax
if
i
if 


 .
)(
.)2/1( 2
2
a
vvvivv
x
ifif
2
)(2)( 2


2ax= vf
2-2vfvi+vi
2+2vivf-2vi
2
vf
2 - vi
2 = 2ax ( explicitly independent of time ).

25. Example : During take off, an airplane travels 960m along a straight runway to
reach a speed of 65m/s before its tires leave the ground. If the motion is
uniformly accelerated, determine
(a) its acceleration,
(b) the elapsed time
(c) its equation of motion
(d) its midway speed.
x=960m
V=0 m/s V= 65 m/s
a= constant midway

26. Solution:
(a) vf
2 - vi
2 = 2ax ;
(65m/s)2 - (0m/s)2 = 2a(960m)
4225 = 1920a
a = 2.2 m/s2
(b) a = (vf -vi) / t
solving for ( t ): t = (vf -vi) / a
t = (65m/s - 0m/s) / ( 2.2m/s2) = 30. sec
(c) x(t) = (1/2)( 2.2m/s2) t2 + (0) t
x(t) = 1.1 t2
(d) vf
2 - vi
2 = 2ax
vf
2 - (0m/s)2 = 2( 2.2 m/s2)[(960/2)m]
vf
2 = 2113
vf = 46 m/s.

27. Freely Falling of Objects:
(1) g = (vf -vi ) / t
(2) y = (1/2) g t2 + vi t
(3) vf
2 - vi
2 = 2 g y

28. Example: A rock is released from a height of 19.6m. Determine
(a) the time it spends in air
(b) its speed just before striking the ground.
19.6m
t=?
vf=?
vi= 0 m/s
(1) g = (vf -vi ) / t
(2) y = (1/2) g t2 + vi t
(3) vf
2 - vi
2 = 2 g y
(+ y)

29. Solution:
Since the rock is not thrown and is only released from rest,
the initial speed, vi = 0.
If the (+ y) axis is taken to be downward, then
g = + 9.8m/s2, and the equation of motion,
y = (1/2) g t2 + vi t
becomes: y = 4.9t2 .
Substituting the given 19.6m for y, we get:
19.6 = 4.9t2
4 = t2 ; t = 2.0s. (The positive root is acceptable).
To find the velocity at which it hits the ground, we may use
vf
2 - vi
2 = 2 g y
vf
2 - 02 = 2 (9.8)(19.6)
vf = 19.6 m/s. (Again the positive answer is acceptable).

30. Example: A rifle fired straight upward at
the ground level and the bullet leaves
the barrel at an initial speed of 315 m/s.
Determine
(a) the highest it reaches
(b) the time to reach the highest point
(c) the return time to the ground, and
its speed just before striking the
ground. Neglect air resistance.
vi= 315 m/s
vf= 0 m/s
h

31. (+ y)

32. Example: A motorcycle ride consists of two segments, as shown in the following
Figure. During segment 1, the motorcycle starts from rest, has an acceleration of
2.6 m/s2, and has a displacement of 120 m. Immediately after segment 1, the
motorcycle enters segment 2 and begins slowing down with an acceleration of -
1.5 m/s2 until its velocity is 12 m/s. What is the displacement of the motorcycle
during segment 2?
a=2.6 m/s2
vi=0 m/s
x=120 m
vf
2 - vi
2 = 2ax
a=-1.5 m/s2
v= 12 m/s
vf1= ? m/s
vf
2 - vi
2 = 2ax
Answer: x2=160 m

33. Example: Using the time and position intervals indicated in the
drawing, obtain the velocities for each segment of the trip.

35. Homework :
problems 1 through 5 chapter 2