1. Motion Along a Straight Line
Chapter 2
Motion Along a Straight Line

2. Motion Along a Straight Line
In this chapter we will study kinematics, i.e., how objects move along a straight line.
The following parameters will be defined:
Displacement Average velocity Average speed Instantaneous velocity Average and instantaneous acceleration
For constant acceleration we will develop the equations that give us the velocity and position at any time.
we will study the motion under the influence of gravity close to the surface of the Earth.
Finally, we will study a graphical integration method that can be used to analyze the motion when the acceleration is not constant.

3. Ch 2-1 Motion Along a Straight Line
Motion of an object along a straight line
Object is point mass
Motion along a horizontal or vertical or inclined (line with finite slope) line
Motion: Change in position
No change in position, body at rest

4. Ch 2-3 Position and Displacement
Axis are used to define position of an object
Position of an object defined with respect to origin of an axis
Position x of an object is its distance from the origin at any time t
Displacement x, a vector, is change in position.
x = xf-xi
When an object changes its position, it has a velocity

5. Check Point 2-1 Ans: x=xf-xi a) x=xf-xi=5-(-3)=+8
b) x=xf-xi=-7-(-3)=-4
c) x=xf-xi=-3-(+7)=-10
Ans: b and c
Here are three pairs of initial and final positions, respectively along x axis. Which pair give a negative displacement?
a) -3 m, + 5 m
b) -3 m, -7 m
c) 7 m, -3 m

6. Ch 2-4 Average Velocity, Average Speed
Average Velocity vavg= x/ t
vavg = (xf-xi) /(tf-ti)=Dispalcemnt/time
Average speed Savg: a scalar
Savg = total distance/ total time
Instantaneous Velocity v:
v= lim (x/ t)
t0
Position-time graph used to define object position at any time t and to calculate its velocity
v is slope of the line on position-time graph

7. Check Point 2-3 Ans: v=dx/dt 1) v=+3 m/s 2) v=-8t m/s 3) v = -4/t3
The following equations give the position x(t) of a particle in four situations ( in each equation , x is in meters, t is in seconds, and t>0):
1) x = 3t-2
2) x=-4t2-2
3) x= 2/t2
4) x=-2
a) In which situation velocity v of the particle constant?
b) In which v is in the negative direction?
Ans: v=dx/dt
1) v=+3 m/s
2) v=-8t m/s
3) v = -4/t3
4) v=0
a) 1 and 4
c) 2 and 3

8. Ch 2-6 Acceleration
When an object changes its velocity, it undergoes an acceleration
Average acceleration aavg
aavg = v/ t
= (vf-vi) /(tf-ti)
Instantaneous acceleration a:
a= lim (v/ t)
t0
= dv/dt=d2x/dt2
Velocity-time graph used to define object velocity at any time t and calculate its acceleration
a is slope of the line on velocity-time graph

9. Ch 2-6 Acceleration
If the sign of the velocity and acceleration of a particle are the same, the speed of particle increases.
If the sign are opposite, the speed decreases.

10. Check Point 2-4
A wombat moves along an x axis. What is the sign of its acceleration if it is moving:
a) in the positive direction with increasing speed,
b) in the positive direction with decreasing speed
c) in the negative direction with increasing speed,
d) in the negative direction with decreasing speed?
Ans:
a) plus
b) minus
c) minus
d) plus

11. Ch 2-7 Constant Acceleration
Motion with constant acceleration has :
Variable Slope of position-time graph
Constant Slope of velocity -time graph
Zero Slope of acceleration -time graph
For constant acceleration
a =aavg= (vf-vi)/(tf-ti)
vavg= (vf+vi)/2

12. Equations for Motion with Constant Acceleration
v=v0+at
x-x0=v0t+(at2)/2
v2=v02+2a(x-x0)
x-x0=t(v+v0)/2
x-x0 =vt-(at2)/2

13. Check Point 2-5
Ans: Table 2-1 deals with constant acceleration case hence calculate acceleration for each equation:
1) a = d2x/dt2=0
2) a = d2x/dt2=-30t+8
3) a = d2x/dt2 = 12/t4-8/t2
4) a = d2x/dt2 = 10
Ans: 1 and 4 ( constant acceleration case)
The following equations give the position x(t) of a particle in four situations:
1) x=3t-4
2) x=-5t3+4t2+6
3) x=2/t2-4/t
4) x=5t2-3
To which of these situations do the equations of Table 2-1 apply?

14. Ch 2-9 Free Fall Acceleration
Free fall acceleration ‘g’ dde to gravity
g directed downward towards Earth’s center along negative y-axis
with a = -g = -9.8 m/s2
equations of motion with constant acceleration are valid for free fall motion

15. Check Point 2-6
What is the sign of the ball’s displacement for the ascent, from the release point to the highest point?
B) What is it for the descent , from the highest point back to to the release point
c) What is the ball’s acceleration at its highest point?
(a) plus (upward displacement on y axis);
(b) minus (downward displacement on y axis);
(c) a = −g = −9.8m/s2

16. Ch 2-10 Graphical Integration in Motion Analysis
Integration of v-t graph to obtain displacement x x =x-x0=vt =v dt
but
v dt= area between velocity curve and time axis from t0 to t
Integration of a-t graph to obtain velocity v Similarly v =v-v0= a dt
a dt = area between acceleration curve and time axis from t0 to t