How Far, or Linear Kinematics
by Chris Murray, November 1997
Table of Contents
Here are the quantities you can know
 s: Displacement
 t: time
 u: Initial velocity
 v: Final velocity
 a: Acceleration
 v_{a}_{v}_{:} Average velocity
These quantities are defined and explained on other pages except for Average velocity which is explained below
New Quantities
And here are the formulas that we have so far:

v_{a}_{v}_{ = }Ds/Dt

a_{a}_{v}_{ = }Dv/Dt

v = u + at
Defining Average velocity
Suppose you knew that a car sped up from 4 m/s to 12 m/s in 3.0 seconds. If I asked
you for the acceleration, at this point, I would expect you to be all over this problem.
But suppose I asked you to calculate the displacement of the car in this time? All we
have so far is v_{a}_{v}_{ = }Ds/Dt
but what velocity do we use? You know that it is going
to be more than 4 m/s x 3.0 s = 12 m, and less than 12 m/s x 3.0 s = 36 m.
How about finding the average velocity of the car during that time? If the acceleration
is uniform, this is valid.
so we have (4 m/s + 12 m/s)/2 = 8 m/s on the average.
The distance then is s = (8 m/s)(3.0 s) = 24 m which is the correct answer.
So we have one new formula:
s = t(u + v)/_{2
}
Which works well as long as the acceleration is constant.
Now we need two more formulas to be fully equipped. Both of them originate from
The formulas we already have:

v_{a}_{v}_{ = }Ds/Dt

a_{a}_{v}_{ = }Dv/Dt

v = u + at

s = t(u + v)/_{2}
And
s = t(u + v)/_{2
}
So, let's start with:
s = t(u + v)/_{2
}
Now let's get rid of v the final velocity substituting (u + at) for v from
The third equation. This gives us:
s = t(u + (u + at))/_{2
}
With a little algebra, this becomes:
s = (2ut + at^{2})/2
s = (2ut+at^{2})/2
And Finally:
s = ut + ^{1}/_{2}at^{2
}
We need another formula, again, let's start with:
s = t(u + v)/_{2
}
This time, let's get rid of time. Solving (v = u + at) for time (t) yields:
t = (v  u)/a
Stuff this into
s = t(u + v)/_{2
}
for t, and you get:
s = ((u + v)/_{2})(v  u)/a
A bit of algebra yields:
v^{2} = u^{2 + }2as
Formulas
So now we have all the formulas we need for solving linear kinematics problems with
uniform acceleration:

v_{a}_{v}_{ = }Ds/Dt

a_{a}_{v}_{ = }Dv/Dt

v = u + at

s = t(u + v)/_{2}

v^{2} = u^{2 + }2as

s = ut + ^{1}/_{2}at^{2}
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 cannon ball reaches a velocity of 324 m/s inside a barrel that is 2.50 m long.
Find a) the time the ball is in the barrel, b) the acceleration of the ball
while it is in the barrel.
Here is what you start with:
s = 2.5 m
t = ?
u = 0 (assumed)
v_{a} = ?
v = 324 m/s
a = ?
So, if you look at the formulas, the fourth one is the ticket, as you know u, v and s, you can solve for t:
so
s = t(u + v)/_{2}
putting in numbers:
2.5 m = t(0 + 324 m/s)/2
so now you know these quantities:

s = 2.5 m

t = .015432 s

u = 0 (assumed)

v_{a} = ?

v = 324 m/s

a = ?
Looking at the formulas, now you can use the this one:
v = u + at
Plug in numbers:
324 m/s = 0 m/s + a(.015432 s)
So a = 324 m/s /(.015432 s) = 20995.2 m/s/s
Now we know everything we want to:
s = 2.5 m
t = .015432 s
u = 0 (assumed)
v_{a} = ?
v = 324 m/s
a = 20995.2 m/s/s
Since the question asked for time and acceleration, you would need to round to the number of significant digits you have. (3 in this case)
so time = .0154 s, and acceleration = 2.10 E4 m/s/s
Go back to: Table of Contents
Example problem 2
A helicopter is ascending at a rate of 12 m/s. When it is 50.0 m above a ship, it
drops an aid parcel intended to land on a deck of the ship. What time does it take to
reach the deck, and with what velocity does it hit the deck?
Here's what we know:
s = 50.0 m (The parcel goes down 50 m. Up is positive)
t = ?
u = +12 m/s (The parcel has the same velocity as the 'copter to start with)
v_{a} = ?
v = ?
a = 9.8 m/s/s
If you look at the formulas:

v_{av = }Ds/Dt

a_{av = }Dv/Dt

v = u + at

s = t(u + v)/_{2}

v^{2} = u^{2 + }2as

s = ut + ^{1}/_{2}at^{2}
The only ones that we can use are the last two. s = ut + ^{1}/_{2}at^{2 }would involve the use of the quadratic equation to solve  so let's see if we can avoid it.
We can use this equation to find the final velocity:
v^{2} = u^{2 + }2as
Plugging numbers into the formula:
v^{2} = (12 m/s)^{2 }+ 2(9.8 m/s/s)(50 m)
So
v^{2} = 144 m^{2}/s^{2} + 980 m^{2}/s^{2}
or
v^{2} = 1124 m^{2}/s^{2}
and finally
v = (1124 m^{2}/s^{2})^{.5} = 33.526 m/s
but, since it is the square root of a square, we really only know that the absolute value is 33.526 m/s. In fact, we know for sure that the final velocity in this case is negative because we made up positive, and the parcel is definitely going down when it hits the boat.
So now we know:
s = 50.0 m (The parcel goes down 50 m. Up is positive)
t = ?
u = +12 m/s (The parcel has the same velocity as the 'copter to start with)
v_{a} = ?
v = 33.5 m/s
a = 9.8 m/s/s
Now you can use the first formula v = u + at to find the time:
33.526 m/s = +12 m/s + (9.8m./s/s)t
so t = (33.526 m/s  12 m/s)/(9.8m./s/s) = 4.64 seconds.
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.
What distance will a train stop in if its initial velocity is 23 m/s and its
acceleration is .25 m/s/s?
(1058 m)
2.
What distance will a car cover accelerating from 12 m/s to 26 m/s in 14 seconds?
(266 m)
3.
A person starts at rest and accelerates at 3.2 m/s/s for 3.0 seconds. What is their
final velocity? What is their average velocity? What distance do they cover in that
time?
(9.6 m/s, 4.8 m/s, 14.4 m)
4.
Steve Apt's group claimed that they fell 3.2 seconds from a cliff into the water.
What was their final speed? How high was the cliff? What is your favorite color?
(31.36 m/s, 50.2 m, Red)
5.
A car going 12.7 m/s accelerates for 14 seconds at 2.6 m/s/s. What is its final
velocity? What distance does it go during that time?
(49.1 m/s, 432.6 m)
Go back to: Table of Contents
6.
What time will it take you to hit the water off of a 10.0 m board? What speed will you
be going when you hit the water?
(1.43 s, 14 m/s)
7.
A car slows from 42 m/s to 21 m/s over a distance of 84 m. What was the
average velocity? What was the time? What was the acceleration?
(31.5 m/s, 2.67 s, 7.9 m/s/s)
8.
A car accelerates from rest down a hill reaching a final speed of 13.7 m/s
over a
distance of 56 m. What was the average speed? What was the time? What was the
acceleration?
(6.85 m/s, 8.2 s, 1.68 m/s/s)
9.
A car skids to a halt in 34 m with an acceleration of 8.2 m/s/s. What was the initial
velocity?
(23.6 m/s)
10.
What must be the acceleration of a train in order for it to stop from 12 m/s in a
distance of 541 m?
(.13 m/s/s)
Go back to: Table of Contents
Solutions to Sample Problems
1.
What distance will a train stop in if its initial
velocity is 23 m/s and its
acceleration is .25 m/s/s?
(1058 m)
Here is what you start with:
 s = ?
 t = ?
 u = 23 m/s
 v_{a} = ?
 v = 0 (it stops)
 a = .25 m/s/s
Well, you can find time from v = u + at:
0 = 23 m/s + (.25 m/s/s)t so t = 92 s
and you can find the displacement from s = t(u + v)/_{2
} = 92*(23 + 0)/2 = 1058 m, the answer.
You also could use v^{2} = u^{2 + }2as and find the answer
in one step: 23^{2} = 0^{2} + 2(.25 m/s/s)s so s = 1058 m
Go to: Problem Formulas Table of Contents
2.
What distance will a car cover accelerating from 12 m/s to 26 m/s in 14 seconds?
(266 m)
Here is what you start with:
 s = ?
 t = 14 s
 u = 12 m/s
 v_{a} = ?
 v = 26 m/s
 a = ?
Find the displacement: s = t(u + v)/_{2
} = 14*(12 + 26)/2 = 266 m
Go to: Problem Formulas Table of Contents
3.
A person starts at rest and accelerates at 3.2 m/s/s for 3.0 seconds. What is their
final velocity? What is their average velocity? What distance do they cover in that
time?
(9.6 m/s, 4.8 m/s, 14.4 m)
Here is what you start with:
 s = ?
 t = 3.0 seconds
 u = 0 (They start at rest)
 v_{a} = ?
 v = ?
 a = 3.2 m/s/s
You can use v = u + at to find the final velocity: v = 0 + (3.2 m/s/s)(3.0 seconds) = 9.6 m/s
Then use v_{a} = (u + v)/_{2} to find the average velocity: v_{a} = (0 + 9.6 m/s)/_{2} = 4.8 m/s
and finally s = v_{a}t to find displacement: s = 4.8 m/s)(3.0 s) = 14.4 m
Go to: Problem Formulas Table of Contents
4.
Steve Apt's group claimed that they fell 3.2 seconds from a cliff into the water.
What was their final speed? How high was the cliff? What is your favorite color?
(31.36 m/s, 50.2 m, Red)
Here is what you start with:
 s = ?
 t = 3.2 seconds
 u = 0 (We assume they don't hurl themselves down the cliff)
 v_{a} = ?
 v = ?
 a = 9.80 m/s/s (The acceleration of free fall on Earth)
Use v = u + at to find the final velocity: v = 0 + (9.80 m/s/s)(3.2 s) = 31.36 m/s
Then try 2as = v^{2}  u^{2} for finding the height of the cliff:
2(9.80 m/s/s)s = 31.36 m/s^{2}  0^{2} = 983.4496 m^{2}/s^{2}
so s = 983.4496 m^{2}/s^{2}/19.6 m/s/s = 50.176 m or about 50.2 m (sf 50. m)
As far as your favorite color, I guess that is a matter of personal choice. I think
that red is a fine personal choice, so I chose it. Don't you think you ought to
choose red too?
Go to: Problem Formulas Table of Contents
5.
A car going 12.7 m/s accelerates for 14 seconds at 2.6 m/s/s. What is its final
velocity? What distance does it go during that time?
(49.1 m/s, 432.6 m)
Here is what you start with:
 s = ?
 t = 14 seconds
 u = 12.7 m/s
 v_{a} = ?
 v = ?
 a = 2.6 m/s/s
Use v = u + at to find the final velocity: v = 12.7 m/s + (2.6 m/s/s)(14 s) = 49.1 m/s
then find the average velocity using v_{a} = (u + v)/_{2}: v_{a} = (12.7 m/s + 49.1 m/s)/_{2} = 30.9 m/s
finally, use s = v_{a}t for the displacement: s = 30.9 m/s)(14 seconds) = 432.6 m or about 430 m with sig figs.
Go to: Problem Formulas Table of Contents
6.
What time will it take you to hit the water off of a 10.0 m board? What speed will you
be going when you hit the water?
(1.43 s, 14 m/s)
Here is what you start with:
 s = 10.0 m (10.0 m boards are that high, not that long)
 t = ?
 u = 0 (assuming falling from rest)
 v_{a} = ?
 v = ?
 a = 9.80 m/s/s (gravity)
to find the time, you need to use s = ut + ^{1}/_{2}at^{2}:
10.0 m = 0t + ^{1}/_{2}(9.80 m/s/s)t^{2}
doing algebra,
2(10.0 m)/(9.80 m/s/s) = t^{2}
so you get an absolute value of time of:
t = 1.4286 s.
then to find the final velocity, the easiest way is to use v = u + at: v = 0 + (9.80 m/s/s)(1.4286 s) = 14 m/s
Go to: Problem Formulas Table of Contents
7.
A car slows from 42 m/s to 21 m/s over a distance of 84 m. What was the
average velocity? What was the time? What was the acceleration?
(31.5 m/s, 2.67 s, 7.9 m/s/s)
Here is what you start with:
 s = 84 m
 t = ?
 u = 42 m/s
 v_{a} = ?
 v = 21 m/s
 a = ?
Find the average velocity with v_{a} = (u + v)/_{2}: v_{a} = (42 m/s + 21 m/s)/_{2} = 31.5 m/s
Then find the time from s = v_{a}t
84 m = (31.5 m/s)t
so t = 84 m / 31.5 m/s = 2.6667 s
and finally, v = u + at will find the acceleration:
21 m/s = 42 m/s + a(2.6667 s)
so
21 m/s = a(2.6667 s), and a = 21 m/s /(2.6667 s) = 7.875 m/s/s (7.9 m/s/s with sf)
Go to: Problem Formulas Table of Contents
8.
A car accelerates from rest down a hill reaching a final speed of 13.7 m/s
over a
distance of 56 m. What was the average speed? What was the time? What was the
acceleration?
(6.85 m/s, 8.2 s, 1.68 m/s/s)
Here is what you start with:
 s = 56 m
 t = ?
 u = 0 (from rest)
 v_{a} = ?
 v = 13.7 m/s
 a = ?
Find the average velocity: v_{a} = (u + v)/_{2} = (0 + 13.7 m/s)/_{2} = 6.85 m/s
Find time: s = v_{a}t plugging in numbers:
56 m = (6.85 m/s)t, so t = 56 m / 6.85 m/s = 8.1752 s
Use v = u + at to find the acceleration:
13.7 m/s = 0 + a(8.1752 s), so a = 13.7 m/s / 8.1752 s = 1.6758 m/s/s (1.7 m/s/s w sf)
Go to: Problem Formulas Table of Contents
9.
A car skids to a halt in 34 m with an acceleration of 8.2 m/s/s. What was the initial
velocity?
(23.6 m/s)
Here is what you start with:
 s = 34 m
 t = ?
 u = ?
 v_{a} = ?
 v = 0 (to a halt)
 a = 8.2 m/s/s (It must be negative if the 34 m is positive)
Police investigating accident scenes solve this problem to determine fault in the case
of excessive speeding. The 34 m would be the length of the skid marks, and the tires
and car would have a characteristic acceleration.
I would use 2as = v^{2}  u^{2} to find the answer in one swell foop:
2(8.2 m/s/s)(34 m) = 0^{2}  u^{2} = u^{2}
getting rid of the minus sign on both sides, and combining on the left side:
557.6 m^{2}/s^{2} = u^{2}, so the absolute value of u must be:
23.61 m/s. Only the positive value has meaning. (This is about 53 mph  Zippy for a
residential neighborhood)
Go to: Problem Formulas Table of Contents
10.
What must be the acceleration of a train in order for it to stop from 12 m/s in a
distance of 541 m?
(.13 m/s/s)
Here is what you start with:
 s = 541 m
 t = ?
 u = 12 m/s
 v_{a} = ?
 v = 0 (stop)
 a = ?
Use 2as = v^{2}  u^{2}:
2a(541 m) = 0^{2}  (12 m/s)^{2} = 144 m^{2}/s^{2}
so
(1082 m)a = 144 m^{2}/s^{2}, a = 144 m^{2}/s^{2}/(1082 m) = .1331 m/s/s or .13 m/s/s
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