Calculus Of One Real Variable – By Pheng Kim Ving Chapter 5: Applications Of The Derivative Part 1 – Section 5.8: Motion 5.8 Motion

 1. Motion On A Line

A car travelling on a straight road can be viewed as a point or an object moving along a line. An apple falling from its tree
can be regarded as an object moving along a line.

Consider an object that is moving along a line, which we assume to be the x-axis. See Fig. 1.1. Recall that the position x
of a point or object on the axis is the signed distance from the origin, whose position is 0, to that point or object. In Fig.
1.1, the position is positive if it's to the right of the origin, and negative if it's to the left. For example,
x > 0, x1 > 0, and
x2 < 0.

Suppose at 1:00:00 PM the object is at position x and at 1:00:05 PM it's at position x1. Clearly the position x of the
object is a function of time
t: x = x(t). Now, 1:00:00 PM is 1 hour or 3600 seconds from mid-day of the same day, and Fig. 1.1   Motion On A Line.

1:00:05 PM is 1 hour 5 seconds or 3605 seconds from that mid-day. Or, 1:00:00 PM is 5400 seconds and 1:00:05 PM is
5405 seconds from 11:30:00 AM of the same day. Or, 1:00:00 PM is 0 second and 1:00:05 PM is 5 seconds from 1:00:00
PM of the same day. In the first case, we choose mid-day as the initial time; in the second and third cases we choose
11:30:00 AM and 1:00:00 PM respectively as the initial time. Of course we can choose one initial time only. 2. Velocity

Recall that the average speed over the time interval [t1, t2] is the average rate of change of distance travelled with
respect to time over [
t1, t2], and that distance travelled is change in position. So (average speed) = (change in position)/
(change in time) = (
x2x1)/(t2t1), where x1 and x2 are positions at times t1 and t2 respectively. As time always
increases, change in time that's non-zero is always positive. Thus, the sign of speed is always the same as that of change
in position. Consequently, if the object moves in the positive direction, then change in position and thus speed are positive,
and if it moves in the negative direction, then change in position and thus speed are negative. The technical term for
signed speed is velocity. The term speed
will be reserved for the magnitude of velocity. Fig. 2.1   Velocity at t is v(t) = x'(t) = dx/dt. The velocity, or instantaneous velocity, v = v(t) of an object at time, or instant, t is defined to be the (instantaneous) rate of change of its position at time t, ie, the derivative of its position x = x(t) at time t: ## Velocity And Direction Of Motion

By Section 5.3 Theorem 2.1 we have:

i.  If v(t) > 0, then the object is moving in the positive direction.

(Intuitively, as dx/dt > 0, a small increase dt > 0 of time produces a small increase dx > 0 of position, eg, from x1 =
1.001 up to
x2 = 1.002 [dx = 0.001], so the object is moving in the positive direction.)

ii.  If v(t) < 0, then the object is moving in the negative direction.
(Intuitively, as
dx/dt < 0, a small increase dt > 0 of time produces a small decrease dx < 0 of position, eg, from x1 =
1.005 down to
x2 = 1.004 [dx = – 0.001], so the object is moving in the negative direction.)

Suppose v(t1) = 0. This means that the object is instantaneously at rest at time t1; it isn't moving in either direction at
that instant
t1. If v(t) = 0 on an interval of time, the object isn't moving for the period over that interval.

 3. Acceleration

A car starting to move from rest is speeding up, ie, its velocity is increasing, perhaps for a few seconds. Velocity, which is
the rate of change of position with respect to time, may itself change with time. That is, it's a function of time. This fact
justifies its functional notation
v(t).

 The acceleration a of the object at time t is the rate of change with respect to time, ie, derivative, a = a(t) of its velocity v = v(t) at time t: Note that since acceleration is the derivative of velocity and velocity is the derivative of position, acceleration is the second
derivative of position. Acceleration that decreases speed is also called deceleration.

## Velocity, Acceleration, And Motion

Remark that: By Section 5.3 Theorem 2.1 and Velocity And Direction Of Motion above we have: If a(t1) = 0, velocity is instantaneously unchanging or stationary at time t1. If a(t) = 0 over a time interval, velocity is
unchanging or constant over that interval.

 4. Finding Velocity And Acceleration

Example 4.1 a.  Find the velocity at t = 3.
b.  Find the acceleration at
t = 3.
c.  At
t = 3, in what direction is the object moving? Is it speeding up or slowing down or going at a constant speed?

Solution c.  Since v(3) > 0 and a(3) < 0, at t = 3 the object is moving in the positive direction and slowing down.

EOS

Remark 4.1 5. Finding Position

Example 5.1

An object accelerates from rest with a constant acceleration of 4 cm/sec2. How far has it travelled by the time its velocity
becomes 16 cm/sec?

Solution When velocity x'(t) = 16, we have 4t = 16, yielding t = 4, and therefore the distance travelled is x(4) = 2 . 42 = 32 cm.
EOS

Remarks 5.1

i.  We think of the object as moving along the x-axis. The choices of the origin and the initial time imply that the initial
values, ie, the values at the initial time, of position, velocity, and acceleration are all 0.

ii.  Velocity and acceleration are given and we're to find position at a particular time, the time when velocity has a

particular value. First we find the equation of position as a function of time t. Next we find the particular time when

velocity has the given value. Then we substitute the value of that particular time into the equation of position.

iii.  Given x''(t) = 4, we antidifferentiate it twice to arrive at the function x(t) = 2t2 by using the conditions x'(t0) = x'(0)
= 0 and
x(t0) = x(0) = 0 to determine the constants C1 and C2 respectively. Since t0 is the initial time and time is
the independent variable of the function
x(t), these conditions are referred to as the initial conditions  for the function

x(t). Now:

y0 = y(0) = 0 + 0 + C2 = C2.

Therefore: where y is in m (meter), g = 9.8 m/sec2, t in sec, v0 in m/sec, and y0 in m. Fig. 6.1   Object falls along y-axis.

Remarks 6.1

i.  Case Where The Object Rises Before It Falls – While the object rises, it moves in the positive direction, so its velocity is
positive. Since the object decelerates because the acceleration g is downward, its speed decreases. Thus, the velocity
decreases. Consequently, the acceleration is again negative.

ii.  Thus, as long as the y-axis is directed upward, the downward acceleration is always negative. The acceleration is
positive, respectively negative, if it's in the same direction as, respectively opposite direction to, the axis of motion.
This fact can be used without proof in solutions to problems. Recall that the situation for velocity is similar. For
example, in the above problem, v0 < 0. If the object rises before it falls and v0 is the initial velocity when the object
starts to rise, then v0 > 0.

iii.  In free-fall problems, it's understood that the y-axis is directed upward and its origin is at the surface of the earth,
unless explicitly stated otherwise.

iv.  Given y''(t) = a = – g, we antidifferentiate it twice to arrive at the function y(t) = – (1/2)gt2 + v0t + y0 by using the
conditions
y'(t0) = v(0) = v0 and y(t0) = y(0) = y0 to determine the constants C1 and C2 respectively. Since t0 is the
initial time and time is the independent variable of the function
y(t), these conditions are referred to as the initial
conditions
for the function y(t). Also see Remarks 5.1 iii.

 Problems & Solutions

1.  A point moves along the x-axis in such a way that its position x at time t is specified by the function x = x(t) = t2
4t + 3.

a.  Find the time intervals on which the point moves to the right.
b.  Find the time intervals on which the point moves to the left.
c.  Find the times at which the velocity is zero.
d.  Find the acceleration when the velocity is zero.
e.  Find the average velocity over the time interval [0, 4].
f.  Find the velocity at time t = 2.
g.  Sketch the graph of x as a function of t.

Solution f.  v(2) = 2(2) – 4 = 0.
g.  2.  A car is travelling at 72 km/h. At a certain instant its brakes are applied to produce a constant deceleration of
0.8 m/sec2. How far does the car travel from the time the brakes are applied to the time it stops?

Solution The initial distance is 0 = x(0) = 0 + 0 + C2 = C2. Hence, x(t) = – 0.4t2 + 20t.

The car stops when velocity x'(t) = 0, ie, when – 0.8t + 20 = 0 or t = 25. It follows that the distance travelled from the
time the brakes are applied to the time the car stops is x(25) = – 0.4(252) + 20(25) = 250 m. 3.  A ball is thrown downward from the top of a 100-m (m = meter) high tower with an initial speed of 2 m/sec. Its height
in meters above the ground t seconds later is y = y(t) = – 4.9t2 – 2t + 100.

a.  How long does it take to reach the ground?
b.  At what instant is its velocity equal to its average velocity?

Solution  4.  A ball is thrown upward from ground level with an initial speed of 35 m/sec so that its height in meters after t seconds
is given by y = y(t) = 35t – 10t2.

a.  How high does the ball go?
b.  How fast does it strike the ground?

Solution

a.  The velocity is v = v(t) = dy/dt = 35 – 20t. The ball is rising when v > 0, ie, when 35 – 20t > 0 or t < 35/20 = 7/4,
and is falling when v < 0, ie, when t > 7/4. So, the ball reaches its maximum height at t = 7/4. Thus, the ball goes to
the maximum height of y(7/4) = 35(7/4) – 10(7/4)2 = 30.625 m.

b.  When the ball is at ground level, we have y = 0, or 35t – 10t2 = 0, so t = 0 or t = 35/10. The ball leaves the ground at
t = 0 and comes back to it at t = 35/10. Consequently, it strikes the ground with velocity v(35/10) = 35 – 20(35/10) =
35 m/sec, hence with speed 35 m/sec. 5.  An object is thrown upward from the roof of a 10-m (m = meter) high building. It rises and then falls back such that
its height above the ground at time t is given by y = y(t) = 10 + 8t – (49/10)t2 until it strikes the ground.

a.  What is the greatest height above the ground that the object attains?
b.  With what speed does the object strike the ground?

Solution

a.  The velocity is v = v(t) = dy/dt = 8 – (49/5)t. The object rises when v > 0, ie, when 8 – (49/5)t > 0 or: hence with a speed of about 16.1247 m/sec.