Linear Acceleration
by Jason "Mussolini" Crandall and Joe "Stalin" Thaler, December 1997
Table of Contents
Here are the quantities you can know:
X Displacement
x_{o} Initial Position
t Time
V_{o} Initial velocity
V Final Velocity
a Acceleration
These quantities are defined and explained on other pages except for Acceleration
which is explained below
New Quantities
And here is the only formula that we have so far:

X = V_{a}t
Defining Acceleration
You already know what velocity is. Velocity is the speed at which an object is
moving at a given. But what if the object is speeding up or slowing down? This is
acceleration. Acceleration is the change in velocity of an object divided by
time. Accleration is measured in meters per second per second(m/s^{2}).
DO NOT confuse velocity and acceleration.
A simple example here will illustrate the difference between acceleration and
velocity. Suppose that a car accelerates from rest to 40 m/s(approximately 90 mi/h)
in 5 s; we're talking supercar here.
To find the acceleration, we must first find the change in velocity. To do this, we
subtract the initial velocity(i.e., 0 m/s) from the final velocity(i.e., 40 m/s). This
calculation gives us a change in velocity of 40 m/s. But, this not the solution to our
problem. We are looking for acceleration and to calculate the acceleration, we must divide
the change in velocity(i.e., 40 m/s) by the amount of time elapsed(i.e., 5 s). We find that
the acceleration of the car during this time is 8 m/s^{2}.
Our new formula is:
a = (V  V_{o})/t
The above example holds true for slowing down, or deceleration, as well. However, the
acceleration will be negative because the final velocity will be less than the initial velocity.
Another common problem is to determine the velocity of an object after a given time at a
given acceleration. That is, if a car accelerates from rest at 2 m/s (something like a
Ford Taurus this time), what its speed after 10 s?
We use the formula we have:
a = (V  V_{o})/t
and we solve for v. First, multiply by t to arrive at:
at = V  V_{o}
We then add V_{o} to each side and we're through.
Our formula is:
V = V_{o} + at
Generally, this is the more useful of the two equations and we will add it to our list of formulas.
Formulas
So now we have all the formulas we need for determining uniform acceleration:

V = V_{o} + at

V_{a} = (V_{o} + V)/_{2}

X = V_{a}t
Go back to: Table of Contents
General Problem Solving Strategy:
 Read the problem.
 Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
 Find any formula that will allow you to calculate
anything that you don't know, and apply it.
 Add what you just found in the last step to your list of knowns.
 Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.
Go back to: Table of Contents
Example problem 1
A train car is moving at +32 m/s when the brakeman applies the brakes, slowing it at a rate
of .75 m/s/s. What time will it take for the train to reach a slower velocity of 12 m/s?
Here is what you start with:
t = ?
V_{o} = 32 m/s
V = 12 m/s
a = .75 m/s/s (this is negative because it is slowing)
So, if you look at the formulas, the first one will work because we are given the initial
and final velocities, as well as the acceleration. The only unknown quantity in this formula
is the time, which is what the question asks for.
So:
V = V_{o} + at
putting in numbers:
12 m/s = (32 m/s) + (.75 m/s/s * t)
so now solve for t:
t = 26.67 s
Now we know everything:
t = 26.67 s
V_{o} = 32 m/s
V = 12 m/s
a = .75 m/s/s
Since the question asks for time, we correct for significant digits (2 in this case) and
arrive at the solution:
time = 27 seconds.
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Example problem 2
A train pulls into the station with a velocity of 5 m/s (Yes, we do have a thing for trains).
It had applied its brakes (which slow it at .35 m/s/s) for 143 seconds prior to that. What
was its velocity before it hit its brakes?
Here's what we know:
t = 143 s
V_{o} = ?
V = 5 m/s
a = .35 m/s/s
If you look at the formulas:

V = V_{o} + at

V_{a} = (V_{o} + V)/_{2}

X = V_{a}t
The only one that we can use is the first(look familiar?).
Plugging numbers into the formula:
5 m/s = V_{o} + (.35 m/s/s * 143 s)
So
V_{o} = 55 m/s
So now we know:
t = 143 s
V_{o} = 55 m/s
V = 5 m/s
a = .35 m/s/s
The problem asks you to find the initial velocity which is 55 m/s so we are through.
Go back to: Table of Contents
Sample Problems
The answers to each problem follow it in parentheses. They also link to a solution to
the problem. Try the problem, check your answer, and go to the solution if you do not
understand.
1.
A car in front of the school goes from rest to 27 m/s in 3.0 seconds. What is its
acceleration?
(+9 m/s/s)
2.
A train can speed up at a rate of .053 m/s/s. What time will it take for it to
reach a speed of 21 m/s from a standing stop?
(396 s)
3.
A rocket accelerates at a rate of 190 m/s/s for 2.4 seconds from rest. What is
its final speed?
(456 m/s)
4.
A car has a velocity of +15 m/s. It then accelerates at a rate of +3.5 m/s/s for
the next 5 seconds. What is its final velocity? What is your favorite ice cream
flavor (chocolate and vanilla do not apply)?
(32.5 m/s, prailines and cream)
5.
A car involved in an 14 m/s collision with a parked car is determined to have skidded
for a time of 3.7 seconds before the impact. If that particular car can brake at
8.2 m/s/s with the tires locked, how fast was the car going before it hit its brakes?
(44 m/s)
Go back to: Table of Contents
6.
A spaceship with a velocity of +320 m/s fires its retro rockets to slow it at a rate
of 22 m/s/s for 3 seconds. How fast is the rocket ship going after that?
(254 m/s)
7.
What time will it take a car that can accelerate from 0 m/s to 20 m/s in 5 seconds to
speed up from 12 m/s to 27 m/s? (Find the acceleration)
(3.75 s)
8.
A baseball can change its velocity from +45 m/s to 52 m/s in the .015 seconds that it
takes to hit a line drive. What is the acceleration of the ball?
(6500 m/s/s Rounded)
9.
What must be your acceleration if you change your velocity from 34 m/s to 21 m/s in
15 seconds?
(.87 m/s/s)
10.
Objects accelerate downwards at 32 f/s/s near the surface of the earth. For how
much time must you fall to reach 60 miles/hour? Convert!
(2.75 s)
Go back to: Table of Contents
Solutions to Sample Problems
1.
A car in front of the school goes from rest to 27 m/s
in 3.0 seconds. What is its acceleration?
(9 m/s/s) Here is what you start with:
t = 3.0 s
V_{o} = 0 m/s (assumed because it starts from rest)
V = 27 m/s
a = ?
Once again, our formula is:
V = V_{o} + at
Plug in the initial and final velocities and the time and calculate the acceleration.
27 m/s = 0 m/s + (a * 3.0 seconds)
Our solution is 9 m/s/s.
Go to: Problem Formulas Table of Contents
2.
A train can speed up at a rate of .053 m/s/s. What time
will it take for it to reach a speed of 21 m/s from a standing stop?
(396 s)
Here is what you start with:
t = ?
V_{o} = 0 (assumed)
V = 21 m/s
a = .053 m/s/s
Once again, our formula is:
V = V_{o} + at
Thus:
21 m/s = 0 + (.053 m/s/s * t)
Solve for t and you get 396 seconds.
Go to: Problem Formulas Table of Contents
3.
A rocket accelerates at a rate of 190 m/s/s for 2.4 seconds
from rest.
What is its final speed? (456 m/s)
Here is what you start with:
t =2.4 s
V_{o} = 0 (assumed)
V = ?
a = 190 m/s/s
You can use V = V_{o} + at to find the final velocity:
V = 0 + (190 m/s/s)(2.4 s) = 456 m/s
Go to: Problem Formulas Table of Contents
4.
A car has a velocity of +15 m/s. It then accelerates at a
rate of +3.5 m/s/s for the next 5 seconds. What is its final velocity? What is your
favorite ice cream flavor (chocolate and vanilla do not apply)?
(32.5 m/s, prailines and cream)
Here is what you start with:
t = 5 s
V_{o} = 15 m/s
V = ?
a = +3.5 m/s/s
Ice Cream = Anything but chocolate and vanilla
Use V = V_{o} + at to find the final Velocity:
V = 15 m/s + (3.5 m/s/s)(5 s) = 32.5 m/s
Now the ice cream part is a little bit more tricky. Without any formulas, we are forced to
rely upon empirical observation. Therefore I took it upon myself to sample all 31 flavors
at BaskinRobbins. One empty wallet and several ice cream headaches later, I have come to
the conclusion that Prailines and Cream is the favorite ice cream flavor. Judging from my
experience, attempts to reproduce this experiment may be hazardous to your health and
financial wellbeing.
Go to: Problem Formulas Table of Contents
5.
A car involved in an 14 m/s collision with a parked car is
determined to have skidded for a time of 3.7 seconds before the impact. If that particular
car can brake at 8.2 m/s/s with the tires locked, how fast was the car going before it hit
its brakes?
(44 m/s)
Here is what you start with:
t = 3.7 s
V_{o} = ?
V = 14 m/s (at impact)
a = 8.2 m/s/s
Use V = V_{o} + at to find the initial Velocity:
14 m/s = V_{o} + (8.2 m/s/s * 3.7 s)
Thus (use a little algebra), V_{o} is 44 m/s. 44 m/s is approximately 98 mph, so our
little friend probably deserves whatever he gets.
Go to: Problem Formulas Table of Contents
6.
A spaceship with a velocity of +320 m/s fires its retro
rockets to slow it at a rate of 22 m/s/s for 3 seconds. How fast is the rocket ship going
after that?
(254 m/s)
Here is what you start with:
t = 3 s
V_{o} = 320 m/s
V = ?
a = 22 m/s/s
Using V = V_{o} + at to find the final velocity:
V = 320 + (22 m/s/s * 3 s) = 254 m/s
Go to: Problem Formulas Table of Contents
7.
What time will it take a car that can accelerate from 0 m/s
to 20 m/s in 5 seconds to speed up from 12 m/s to 27 m/s? (Find the acceleration)
(3.75 s)
Here we have two problems:
t = ?
V_{o} = 12 m/s
V = 27 m/s
a = ?
This is not yet solvable, because we don't know the acceleration.
Let's assume that the acceleration from 12 m/s to 27 m/s is the same
as it is from 0 to 20 m/s. In order to find
the acceleration we use our
favorite formula again: V = V_{o} + at.
20 m/s = 0 m/s + (a * 5 s) (Look at the first 3 numbers
and we get a = 4 m/s/s
We now know:
t = ?
V_{o} = 12 m/s
V = 27 m/s
a = 4 m/s/s
Now find the time using the our favorite formula once again: V = V_{o} + at.
27 m/s = 12 m/s + (4 m/s/s * t)
Thus, solving for t, we arrive at the answer: 3.75 seconds.
Go to: Problem Formulas Table of Contents
8.
A baseball can change its velocity from +45 m/s to 52 m/s
in the .015 seconds that it takes to hit a line drive. What is the acceleration of the ball?
(6500 m/s/s Rounded)
Here is what you start with:
t = .015 s
V_{o} = +45 m/s
V = 52 m/s
a = ?
Find the acceleration using the our freindly neighborhood aceleration formula once again:
V = V_{o} + at.
Plug in a couple of numbers and get:
52 m/s = 45 m/s + (a * .015 s)
Solve for a, and we get approximately 6466.66 m/s/s, which, in case you didn't know, is
really, REALLY fast. Using sig figs, we get 6500 m/s/s.
Go to: Problem Formulas Table of Contents
9.
What must be your acceleration if you change your velocity
from 34 m/s to 21 m/s in 15 seconds?
(.87 m/s/s)
Here is what you start with:
t = 15 s
V_{o} = 34 m/s
V = 21 m/s
a = ?
Okay boys and girs, let's see if we can possibly guess what formula we're going to use today.
What's that I hear you saying? Very good! It IS: V = V_{o} + at.
Let's go to the numbers:
21 m/s = 34 m/s + (a * 15 s)
After a random survey, the majority of people polled feel that .87 m/s/s is the correct answer.
However, 9% of our respondents DO NOT like Prailines and Cream.
Go to: Problem Formulas Table of Contents
10.
Objects accelerate downwards at 32 f/s/s near the surface
of the earth. For how much time must you fall to reach 60 miles/hour? Convert! (2.75 s)
Here is what you start with:
t =
V_{o} = 0 (assumed)
V = 60 miles per hour = 88 f/s
a = 32 f/s/s
Okay, I'm getting kind of tired of restating our favorite formula, so here it is with the
numbers already plugged in:
88 f/s = 0 + (32 f/s/s * t)
Me: I'll take acceleration for $500 Alex.
Alex: And the answer is, 88 f/s divided by 32 f/s/s.
Me: What is 2.75 seconds?
Alex: That's right! You win, now get off the web and go do something useful!
Go to: Problem Formulas Table of Contents