Calculus
Of One Real Variable – By Pheng Kim Ving 
5.8 
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& Solutions
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 xaxis. 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, x_{1} > 0, and
x_{2} < 0.
Suppose at 1:00:00 PM the object is at
position x and at 1:00:05 PM it's at position x_{1}. 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
midday 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 midday. 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 midday 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.
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2. Velocity 
Recall that the average speed over the time interval [t_{1}, t_{2}] is the average rate of change of distance travelled with
respect to time over [t_{1}, t_{2}], and that distance travelled is change in
position. So (average speed) = (change in position)/
(change in time) = (x_{2} – x_{1})/(t_{2} – t_{1}), where x_{1} and x_{2} are positions at times t_{1} and t_{2} respectively. As time always
increases, change in time that's nonzero 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)

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 x_{1} =
1.001 up to x_{2} = 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 x_{1} =
1.005 down to x_{2} = 1.004 [dx = – 0.001], so
the object is moving in the negative direction.)
Suppose v(t_{1}) = 0. This means
that the object is instantaneously at rest at time t_{1}; it isn't moving
in either direction at
that instant t_{1}. If v(t) = 0 on an
interval of time, the object isn't moving for the period over that interval.
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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.
Remark that:
By Section 5.3 Theorem 2.1 and Velocity And Direction Of Motion above we have:
If a(t_{1}) = 0, velocity is instantaneously unchanging
or stationary at time t_{1}. If a(t) = 0 over a time interval, velocity is
unchanging or constant over that interval.
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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
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5. Finding Position 
Example 5.1
An object accelerates from rest
with a constant acceleration of 4 cm/sec^{2}. 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 . 4^{2} = 32 cm.
EOS
Remarks 5.1
i. We think of the object as moving along the xaxis. 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) = 2t^{2} by using the
conditions x'(t_{0}) = x'(0)
= 0 and x(t_{0}) = x(0) = 0 to determine the constants C_{1} and C_{2} respectively. Since t_{0} 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).
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6. General Equation For Free Fall 
Now:
y_{0} = y(0) = 0 + 0 + C_{2} = C_{2}.
Therefore:

where y is in m (meter), g = 9.8 m/sec^{2}, t in sec, v_{0} in m/sec, and y_{0} in m.
Fig. 6.1 Object falls along yaxis. 
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 yaxis
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, v_{0} < 0. If the
object rises before it falls and v_{0} is the initial
velocity when the object
starts
to rise, then v_{0} > 0.
iii. In freefall problems, it's understood that
the yaxis 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)gt^{2} + v_{0}t + y_{0} by using the
conditions y'(t_{0}) = v(0) = v_{0} and y(t_{0}) = y(0) = y_{0} to determine the constants C_{1} and C_{2} respectively.
Since t_{0} 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 xaxis in such a
way that its position x
at time t
is specified by the function x
= x(t) = t^{2} –
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/sec^{2}. 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 + C_{2} = C_{2}. Hence, x(t) = – 0.4t^{2} + 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(25^{2})
+ 20(25) = 250 m.
3. A ball is thrown downward from the top of a
100m (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) = – 4.9t^{2} – 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 – 10t^{2}.
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 – 10t^{2} = 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 10m (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)t^{2} 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.
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