#### Motion

In This Chapter:✔

*V*

*elocity*

✔

*Acceleration*

✔

*Distance, Velocity, and Acceleration*

✔

*Acceleration of Gravity*

✔

*Falling Bodies*

✔

*Projectile Motion*

*Velocity*The

*velocity*of a body is a vector quantity that describes both how fast it is moving and the direction in which it is headed.

In the case of a body traveling in a straight line, its velocity is simply the rate at which it covers distance. The

*average velocity ¯v*of such a body when it covers the distance

*s*in the time

*t*is

*v*=

*s/*

*t*

Average velocity = distance/time

The average velocity of a body during the time *t*does not completely describe its motion, however, because during the time

*t*, it may sometimes have gone faster than

*¯v*and sometimes slower. The velocity of a body at any given moment is called its

*instantaneous velocity*and is given by

Here, D

*s*is the distance the body has gone in the very short time interval D

*t*at the speciﬁed moment. (D is the capital Greek letter

*delta*.) Instanta- neous velocity is what a car’s speedometer indicates.

When the instantaneous velocity of a body does not change, it is moving at

*constant velocity*. For the case of constant velocity, the basic formula is

*s*=

*v*

*t*

Distance = (constant velocity)(time)

**Solved Problem 2**.

**1**The velocity of sound in air at sea level is about 343 m/s. If a person hears a clap of thunder 3.00 s after seeing a lightning ﬂash, how far away was the lightning?

**Solution**. The velocity of light is so great compared with the velocity of sound that the time needed for the light of the ﬂash to reach the person can be neglected. Hence

*s*=

*vt*= (343 m/s) (3.00 s) = 1029 m = 1.03 km

######
*Acceleration*

A body whose velocity is changing is accelerated. A body is accelerated when its velocity is increasing, de- creasing, or changing its direction.*Acceleration*

The

*acceleration*of a body is the rate at which its velocity is changing. If a body moving in a straight line has a velocity of

*v*0 at the start of a certain time interval

*t*and of

*v*at the end, its acceleration is

A positive acceleration means an increase in velocity; a negative acceleration (sometimes called

*deceleration*) means a decrease in velocity. Only constant accelerations are considered here.

The deﬁning formula for acceleration can be rewritten to give the ﬁnal velocity

*v*of an accelerated body:

*v*=

*v*0 +

*at*

Final velocity = initial velocity + (acceleration)(time)

We can also solve for the time *t*in terms of

*v*0,

*v*, and

*a*:

Velocity has the dimensions of distance/time. Acceleration has the dimensions of velocity/time or distance/time2. A typical acceleration unit is the meter/second2 (meter per second squared). Sometimes two different time units are convenient; for instance, the acceleration of a car that goes from rest to 90 km/h in 10 s might be expressed as

*a*= 9 (km/h)/s.

**Solved Problem 2**.

**2**A car starts from rest and reaches a ﬁnal velocity of 40 m/s in 10 s. (

*a*) What is its acceleration? (

*b*) If its acceleration remains the same, what will its velocity be 5 s later?

**Solution**. (

*a*) Here

*v*

_{0}= 0. Hence

(

*b*) Now

*v*0 = 40 m/s, soDistance, Velocity, and Acceleration

Let us consider a body whose velocity is

*v*0 when it starts to be accelerated at a constant rate. After time

*t*, the ﬁnal velocity of the body will be

*v*=

*v*

_{0}+

*at*

*t*? The average velocity

*v¯*of the body is

and so

Since

*v*=

*v*0 +

*at*, another way to specify the distance covered during

*t*is

If the body is accelerated from rest,

*v*0 = 0 and

Another useful formula gives the ﬁnal velocity of a body in terms of its initial velocity, its acceleration, and the distance it has traveled during the acceleration:

This can be solved for the distance

*s*to give

In the case of a body that starts from rest,

*v*

_{o}= 0 and

Table 2.1 summarizes the formulas for motion under constant acceleration.

__Acceleration of Gravity__All bodies in free fall near the earth’s surface have the same downward acceleration of

*g*= 9.8 m/s2 = 32 ft/s

^{2}

**Y****ou Need to Know**###### A body in free fall has the same downward acceleration whether it starts from rest or has an initial velocity in some direction.

The presence of air affects the motion of falling bodies partly through buoyancy and partly through air resistance. Thus two different objects falling in air from the same height will not, in general, reach the ground at exactly the same time. Because air resistanceincreases with velocity, eventually a falling body reaches a

*terminal velocity*that depends on its mass, size, and shape, and it cannot fall any faster than that.

__Falling Bodies__When buoyancy and air resistance can be neglected, a falling body has the constant acceleration g and the formulas for uniformly accelerated motion apply. Thus a body dropped from rest has the velocity

*v*=

*gt*

*t*, and it has fallen through a vertical distance ofFrom the latter formula, we see that

and so the velocity of the body is related to the distance it has fallen byTo reach a certain height

*h*, a body thrown upward must have the same initial velocity as the ﬁnal velocity of a body falling from that height, namely,

*v*= 2

*gh*.

**Solved Problem 2**.

**3**What velocity must a ball have when thrown up- ward if it is to reach a height of 15 m?

**Solution**. The upward velocity the ball must have is the same as the downward velocity the ball would have if dropped from that height. Henc

*Projectile Motion*The formulas for straight-line motion can be used to analyze the hori- zontal and vertical aspects of a projectile’s ﬂight separately because these are independent of each other. If air resistance is neglected, the horizon- tal velocity component

*v*

*x*remains constant during the ﬂight. The effect of gravity on the vertical component

*v*

*y*is to provide a downward accel- eration. If

*v*

*y*is initially upward,

*v*

*y*ﬁrst decreases to 0 and then increases in the downward direction.

The range of a projectile launched at an angle

*q*above the horizon- tal with initial velocity

*v*0 isThe time of ﬂight iIf θ

_{1}is an angle other than 45° that corresponds to a range

*R*, then a second angle θ

_{2}for the same range is given by

θ

as shown in Figure 2−1._{2 }= 90° − θ_{1}
Figure 2-1

**Solved Problem 2**.

**4**A football is thrown with a velocity of 10 m/s at an angle of 30° above the horizontal. (a) How far away should its intended receiver be? (b) What will the time of ﬂight be?

**Solution**.