Calculus using
Distance, Velocity and
Acceleration
By Tony Farah
Distance (Displacement)
 Distance can be defined as
the difference between point a
and point b as a function of
time
 An example would be that a
baseball is hit and falls just
shy of homerun
 The displacement, or its
distance above the ground, is
from the time the ball hits the
bat until the ball falls into
center field
 It can be defined in the
following function where y is
displacement in feet above
the ground and t is the
number of seconds since the
ball was hit
 Distance is measured in feet
y (t) = -14t2 + 23t + 2
Velocity
y (t) = -14t2 + 23t + 2
y ’ (t) = -28t + 23
 Velocity can be defined
as how fast an object is
going and in what
direction
 Velocity can also be
defined as an
instantaneous rate of
change or a derivative
 In this case the velocity
is how fast the ball is hit
and in what direction it
travels
 Velocity = dy/dt or y ’ (t)
(y prime of t)
 Velocity is in ft/sec
Acceleration
 Acceleration occurs
when velocity changes
 It is the derivative of
velocity and the 2nd
derivative of
displacement
 Negative acceleration
means velocity is
decreasing
 It can be defined as
dv/dt or y ’’ (t) (y double
prime of t)
 Acceleration is in
ft/sec/sec
y (t) = -14t2 + 23t + 2
y ’ (t) = -28t + 23
y ” (t) = -28
Key Information
 To find velocity at a particular time, substitute for t in y
(t)
 Negative velocity means distance is decreasing
 If velocity and acceleration have the same sign, speed
increases and vice versa
 Negative acceleration means velocity is decreasing
 Distance is measured in ft, velocity in ft/sec, and
acceleration in ft/sec/sec
 A change in direction can be indicated when a graph
goes from negative to positive values or from positive
to negative values
Example Number 1
Jane and some friends are playing kickball
after school. Jane kicks the ball so hard it
flies up and over the fence. As it rises and
falls its distance above the ground is a
function of time since the kick measured in feet.
The function can be defined as f (t) = -30t3 + 12t2 + 8t +2
a)
b)
c)
Find velocity and determine what velocity is at t = 1
Is the distance increasing or decreasing at t = 1 and
explain
Determine acceleration of the function and determine
acceleration at t=1
Part A
Velocity is the derivative of acceleration so
take the function’s derivative
f (t) = -30t3 + 12t2 + 8t +2
f ’ (t) = -90t2 + 24t + 8
Now since we know velocity we can plug
into the equation for t.
f’ (1)= -90(1)2 + 24(1)+ 8
f’ (1)= -58 feet/sec
Part B
Now since we know velocity = -58 ft/sec we can
determine what this means in terms of
distance. The velocity here is decreasing
because the its sign is negative at t = 1.
Whenever velocity is equals a negative
number its distance is always decreasing. From
this we can also concur that positive velocity
means that distance is increasing.
Part C
We can find acceleration by just taking the
derivative of velocity.
f ’ (t) = -90t2 + 24t + 8
f ” (t) = -180t + 24
Now we need to find acceleration at t = 1
f ” (1) = -180(1) + 24
f ” (1) = -156 ft/sec/sec
Example Number 2
Find the average
velocity at t = 3 by
using the average rate
of change
Hint: y2 – y1
x2 – x 1
t
d (t)
0
10
2
4
4
17
6
21
How To Solve It
To use this the formula we can pick any
two y values and any two x values and find
its slope or average rate of change. The best
estimate would be to use the values closest to
t=3
17 – 4
4–2
= 13
2
This answer is the average velocity at t = 3
Multiple Choice # 1
(No Calculator)
A bug begins to crawl up a vertical wire at time t = 0. The velocity v of the
bug at time t, 0 < t < 8, is given by the function whose graph is shown above.
At what value of t does the bug change direction?
a.) 2
b.) 4
c.) 6
d.) 7
e.) 8
Answer key to Multiple
Choice # 1
Answer = c.) 6
The reason for this is that the graph goes
or changes from positive to negative values
at t = 6 indicating a change in direction.
Multiple Choice # 2
(with calculator)
s (t) = t2 - 20
Find the average velocity from t = 3 to t = 5
a.) 2
b.)4
c.)6
d.)8
Answer Key to Multiple
Choice #2
For this problem we must use the average
velocity formula
s(5) – s(3) = 5 – (-11) = 16 =
5–3
2
2
Choice d.) 8
8
Multiple Choice # 3
(no calculator)
An objects distance from its starting point
at time (t) is given by the equation
d (t) = t3 - 6t2 - 4. What is the speed of
the object when its acceleration is 0?
a.) 2
b.) -24
c.) 22
d.) 44
e.) -12
Answer Key to Multiple
Choice # 3
Choice
a.) 2
You must take the derivative of distance
then the derivative of velocity
d’ (t) = 3t2 - 12t
d” (t) = 6t – 12
Now you must solve for t by setting d” or
acceleration = 0
6t – 12 = 0
+12 +12
=
6t = 12
6
6
t=2
Multiple Choice # 4
A particle moves along the x-axis so that its
position at time t is given by
d (t) = t2 – 6t + 5. For what value of t is the
velocity of the particle zero?
a.) 1
b.) 2
c.) 3
d.) 4
e.) 5
Answer Key to Multiple
choice # 4
Choice a.) 3
First we find the derivative of the distance
than we set = 0 just like multiple choice
problem # 3
V(t) = 2t – 6
2t - 6 = 0
+6 +6
2t = 6
2 2
t=3
Free Response # 1 (you
may use your calculator)
Professor Frink never anticipated Truckasaurus getting loose, but then again, he never
foresaw the obvious dangers in reanimating the corpse of his long-dead father. Frinkie's
not one for long-term planning. Anyway, Homer Simpson lies directly in the path of
the flame-spewing juggernaut, with only the meager acceleration of the family station
wagon standing between him and utter destruction.
Assume Homer's velocity (in feet per second) is given by the equation , where t is
measured in seconds and . Answer the following questions based on the given
information, accurate to the thousandths place.
(a) Is Homer traveling backwards at any time during the first 7 seconds of his escape
attempt? If so, during what time interval(s) is the gear shift in reverse?
(b) What is Homer's average velocity from t = 2 to t = 5?
(c) At what time(s) does Homer change from accelerating to decelerating, or vice versa?
Answer to Free Response
#1
(a) Homer is traveling backward whenever his velocity is negative. In other words,
you're looking for the intervals of v(t) when its graph falls below the x-axis. If you
graph v on your calculator, you'll find 2 x-intercepts in the interval [0,7]. It's
advisable to use the calculator to find them: x = .40642065 and x = 4.1819433. Since
the graph is below the x-axis between those x-intercepts, Homer's in reverse on the
interval (.406, 4.182).
(b) Since you're given the velocity function, and are asked to find its average value, you
have to use the average value formula, which states that the average value of f on
the interval [a,b] is . For this problem, a = 2 and b = 5. Since you're allowed to use a
calculator, it would be a waste of time to compute the antiderivative by hand. The
answer is -3.75.
(c) Homer will change from acceleration to deceleration (or vice versa) whenever the
derivative of f has an x-intercept (since the derivative of the velocity function is the
acceleration function). (Note that f also must cross the axis there, as well, not simply
bounce off of it, or there will be no change in the sign of the graph.) The derivative
of f is . The acceleration function has only one x-intercept on the interval [0,7]: x =
2.786 seconds, at which Homer goes from decelerating to accelerating.
Free Response # 2
A particle moves along the y-axis with the velocity given
by v (t) = t sin ( t2 ) for t is greater than or equal to 0.
a.) In which direction (up or down) is the particle moving
at time t = 1.5? Why?
b.) Find the acceleration of the particle at time t = 1.5. Is
the velocity of the particle increasing at t = 1.5? Why or
why not?
c.) Given that y (t) is the position of the particle at time t
and that y (0) = 3, find y (2).
d.) Find the total distance traveled by the particle from t =
0 to t = 2.
Answer Key to Free
Response # 2
a.) v (1.5) = 1.5sin(1.52) = 1.167
Up, because v (1.5) is greater than 0
b.) a (t) = v’ (t) = sin t2 + 2 t2 cos t2
a (1.5) = v’ (1.5) = -2.048 or -2.049
No; v is decreasing at 1.5 because v’ (1.5)
is less than 0
Continued Answer Key
c.) y(t) = the integral of v(t) dt
= the integral of tsin t2 dt = - cos t2 + C
2
y (0) = 3 = -1+ C = 7
2
2
y (t) = - 1cos t2 + 7
2
2
y (2) = - 1cos4 + 7 = 3.826 or 3.827
2
2
d.) distance = the integral from 0 to 2 of the
absolute value of v(t) = 1.173
Works Cited
Mr. Watson Class Notes and Worksheets
“Average velocity.” Chapter 20-4 Velocity and
Acceleration. 7 June 2007
<http://home.alltel.net/okrebs/page205.html>.
Foerster, Paul A. Calculus. EmeryVille: Steve
Rasmussen, 1998.
“Problem of the Year 2003-2004.” Calculus-Help. 7 June
2007 <http://www.calculus-help.com/>.