Lab 4
Math 111 Spring 2016
Name: ___________________________________________________ Section: _____ Score: ______
The Kinematics of Linear Motion
In this lab you will learn how to use calculus to compute some functions which
represent 1 the kinematics of linear motion.
Velocity and Acceleration: Imagine an object that is located at the origin and can only
travel along the s-axis. Its position along that line is given by a. function, s(t), where t
represents time and s(t) gives the x coordinate position of the object at time t. Since the
object is moving, it has a velocity, denoted by v(t).
The velocity of the object is the first derivative of the position function, s(t). Velocity
will be positive when the object is moving to the right, zero in the instant when it is
pausing to change direction, and negative when it is moving to the left.
The second derivative of a function s(t) is found by taking the derivative twice. The
second derivative of s(t) (thus the first derivative of v(t)) is called the acceleration, and
denoted a(t). It will be positive when the object's speed is increasing and negative when
it is decreasing. We use the word accelerate in exactly that way when we refer to the
speed of our car when we accelerate, we are increasing our speed. Deceleration, i.e.,
going slower, corresponds to a negative value of acceleration.
Summary:
Position/Displacement
s(t)
Velocity
vt   s t 
Acceleration
at   vt   st 
1
Example:
s(t) = t2
(units are feet)
v(t) = 2t
(units are ft/sec)
a(t) = 2
(units are ft per sec per sec)
The word "kinematics" refers to those aspects of motion that arise from the use of calculus
to connect displacement, velocity and acceleration, By contrast, one studies the "dynamics" of
bodies in motion by linking forces and characteristics of the bodies to the kinematic properties
of that motion. Dynamical problems are usually done in physics courses. One does not need to
know anything about dynamics to work kinematic problems.
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The relationship between velocity and acceleration can be tricky to understand. Let’s look at two
cases where the particle or object is moving to the right (thus the velocity is positive):
 acceleration is positive (velocity is increasing), then the object is “speeding up” while moving
to the right.
 acceleration is negative (velocity is decreasing), then the object is “slowing down” while
moving to the right. These cases should be somewhat intuitive since they parallel your
experience in a car that is moving forward.
Now let’s look at the two cases where the velocity is negative (the object is moving to the left).
 acceleration is positive (velocity is increasing, which in this case means velocity is becoming
less negative), then the object is “slowing down” while moving to the left.
 acceleration is negative (velocity is decreasing, which in this case means the velocity is
becoming more negative), then the object is “speeding up” while moving to the left.
Let’s summarize this in a table for the case of the object that moves along a straight line.
Summary:
Acceleration Positive +
Acceleration Negative -
Velocity Positive
+
Object moving to the right and
“speeding up”
Object moving to the right and
“slowing down”
Velocity Negative
Object moving to the left and
“slowing down”
Object moving to the left and
“speeding up”
Visualizing the path of an object on a straight line path
Study the diagram of the path of an object below. The object is really sliding back and forth
on a straight track but we can’t see that motion well unless we open up the path this way. When the
stopwatch starts timing at t = 0 seconds, the object is 4 feet on the displacement scale.
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Problem 1: Use the diagram displayed on the previous page. In the questions below, “distance” is
that used to compute gasoline mileage or wear and tear on the vehicle. Displacements forward or in
reverse both contribute.
a) What is the distance traveled in the first 3 seconds?
b) What is the distance traveled in the first 5 seconds?
c) What is the distance traveled in the first 8 seconds?
d) When is the object at rest?
Problem 2: An object moves along a line according to a law of motion s = f(t), where t is measured
in seconds and s in feet. For f(t) = t3 – 9t2 + 15t + 10.
a) Find the velocity at time t.
b) What is the velocity after 3 seconds?
c) When is the object at rest? (Hint: It is at rest when the velocity is 0.)
d) When is the object moving in the positive direction?
e) Find the distance traveled in the first 8 seconds.
f) Draw a diagram like that used in Problem 1 to illustrate the motion of the object. (Hint: Be careful
to start your path at the correct point on the number line.)
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Now add several more parts to the problem, recalling that acceleration is the derivative of
velocity.
g) Find the acceleration of the object at time t.
h) Find the acceleration after 2 seconds.
i) When is the acceleration zero?
j) When is the acceleration positive?
Notice that since f(t) was a cubic function, we can expect its velocity to be a _____________
function and its acceleration to be a ________________ function.
Problem 3: Now we are going to look at a motion problem where a ball is moving vertically under
the influence of gravity, a force taken as constant near the surface of the earth. In this instance the
position function represents the height of the ball above the ground, and is given by a quadratic
function. Think about the following and fill in the blanks with the word positive, negative or zero.
The answers are in a box at the bottom of the last page of this lab.
a) On the way up the velocity of the ball is ___________________.
b) At the top of its excursion, the velocity of the ball is ___________________.
c) On the way down the velocity of the ball is ___________________.
d) On the way up the acceleration of the ball is ___________________ (Think: velocity is
decreasing).
e) On the way down the acceleration of the ball is ___________________ (Think: velocity is getting
more negative).
Now suppose the position function, i.e. the height the ball is above the ground, in this problem
is given by2 s(t) = 24t – 16t2 where s is given in feet and t in seconds.
f) The velocity of the ball at time t is given by the function v(t) = ___________________.
2
The 16 in this formula is a characteristic of the earth. It is slightly accurate to quote the
number as 16.1. The 24 is determined by the individual who throws the ball.
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g) The acceleration of the ball at time t is given by the function ___________________.
h) Determine when the ball reaches its maximum height, which is the time when the velocity is 0.
___________________.
i) What is the maximum height reached by the ball? ___________________
j) Graph the function s(t). Sketch the graph below. Label the axes and intercepts.
k) What is the velocity of the ball when it starts its journey from the ground (t = 0)?
l) What is the velocity of the ball 8 feet from the ground and on its way down?
m) What is the acceleration of the ball at time t = 0? ___________________;
at t = 1/2? ___________________; at time t? ___________________.
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Problem 4: Show all your work.
If a ball is thrown vertically upward with a velocity of 80 ft/sec, then its height after t seconds is
modeled by the equation s t   80t  16t 2 .
(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 96 feet above the ground on its way up?
(c) What is the velocity of the ball when it is 96 feet above the ground on its way down?
The velocity is positive on the way up and negative on the way down.
It is zero when the object reaches the highest point. The acceleration
is always negative in this case (while moving upward, the velocity is
decreasing; while moving downward, the velocity is getting more
negative).
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