How Far II : The Sequel
by Dustin Glazier, January 1998
Table of Contents
Basically, this page is for more practice of your astounding Linear Kinematic skills
So, if something looks unfamiliar, than try going back to one of the previous pages.
Moving right along,
Here's the quantities you can know:
X Displacement
t Time
V_{o} Initial velocity
V_{a} Average Velocity
V Final Velocity
a Acceleration
These quantities are defined and explained on other pages
Formulas
So we already have all the formulas we need for solving linear kinematics problems with
uniform acceleration, but since I am here to help, here they are again:

V = V_{o} + at

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

X = V_{a}t

X = V_{o}t + ^{1}/_{2}at^{2}

2aX = V^{2}  V_{o}^{2}, Which I like to have setup as: V_{f}^{2} = V_{o}^{2} + 2aX
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
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. (Don't worry, you wouldn't be the first!)
1.
A baseball leaves the bat with an upward velocity of 54 m/s.
What time does it take to reach the top? How high does it go?
What total time will it be in the air?
(149 m, 5.5 s, 11s)
2.
A person jumps off of a cliff and hits the water below moving
with a velocity of 24 m/s. What time were they in the air?
How high is the cliff?
(2.4 s, 29 m)
3.
Cliff divers in South America jump from 300 foot cliffs into
the water. (1 m = 3.281 f) What time does it take them to hit the water,
and how fast are they going
when they do hit the water?
(4.3 s, 42 m/s)
4.
Red Elk leaves the 10.0 m diving board with an
upward velocity and hits the water 1.9 seconds later. What was his initial upward velocity?
To what height above the diving board did he rise before going down?
With what speed did he hit the water?
(4.0 m/s, .84 m, 14.6 m/s)
5.
A car will skid to a halt at a rate of 9.4 m/s/s.
If you measure skid marks that are 34 m long, with what speed
was the car going that made them?
(25 m/s)
Go back to: Table of Contents
6.
A train can speed up at .15 m/s/s. In what minimum
distance can it attain a speed of 25 m/s starting from rest?
(2083 m)
7.
A drag racer can reach a speed of 53 m/s over
a distance of 120 m. What is its acceleration? Over what distance
can it reach a speed of 85 m/s?
(11.7 m/s/s, 309 m)
8.
A jetliner must reach a speed of 80 m/s to take off,
and can accelerate at 4.7 m/s/s. What is the minimum length of runway?
(681 m)
9.
Theoretically, what would be the velocity of a steel marble dropped from an airplane
1000 m above the ground just as it hits the ground?
(140 m/s)
10.
A rifle bullet leaves the muzzle of a .75 m long barrel going 450 m/s.
What is the acceleration of the bullet while it is in the barrel?
(135,000 m/s/s)
Go back to: Table of Contents
Solutions to Sample Problems
1.
A baseball leaves the bat with an upward velocity of 54 m/s.
How high does it go? What time does it take to reach the top?
What total time will it be in the air?
(149 m, 5.5 s, 11s)
Here is what you start with:
 X = ? (we will call the displacement on the yaxis X)
 t = ?
 V_{o} = 54 m/s
 V_{a} = ?
 V = 0 (it hits the ground)
 a = 9.8 m/s/s This is gravity
Well let's find an equation that solves for X
Of the many that do fit, 2aX = V^{2}  V_{o}^{2} is best because we know all the variables exept for X.
Plug in those cute numbers:
2(9.8 m/s^{2})X = (0)^{2}  (54 m/s)^{2}
X = 2916 m/s

19.6 m/s^{2}
X = 148.77 which is only 149m with sig figs
We now move on to the time it takes for it to reach the top:
I think that X = V_{o}t + ^{1}/_{2}at^{2} looks appealing so let's try it.
We then plug in the numbers:
149 = 0t + ^{1}/_{2}(9.8 m/s^{2})t^{2}
Finishing it through:
149 = 4.9t^{2}
Leaving us with t^{2} = 30.40816327
take the square root of that and it gives you t = 5.5 seconds
And to find the total time in the air we simply double the time to get to the top>
Because the acceleration is the same for both going up and down, so is the time to reach
Therefore the time for the whole flight is t = 11 seconds
Go to: Problem Formulas Table of Contents
2.
A person jumps off of a cliff and hits the water below moving
with a velocity of 24 m/s. What time were they in the air?
How high is the cliff?
(2.4 s, 29 m)
Here is what you start with:
 X = ?
 t = ?
 V_{o} = 0 m/s
 V_{a} = ?
 V = 24 m/s
 a = 9.8
Let's plug this in to the formula that we know all the variables for, except for t
V = V_{o} + at, seems the most appropriate
So, 24 m/s = (0) + (9.8m/s^{2})t
Therefore we divide 24 by 9.8 and it gives us t = 2.448979592 seconds
With this we can now find the height of the cliff
With X = V_{a}t
V_{a} = 12 m/s because V_{a} = (V_{o} + V)/_{2}
and X = (12 m/s)(2.4 s)
Therefore X = 29m but we know that the negative sign denotes that the jumper went down.
Go to: Problem Formulas Table of Contents
3.
Cliff divers in South America jump from 300 foot cliffs into
the water. (1 m = 3.281 f) What time does it take them to hit the water,
and how fast are they going
when they do hit the water?
(4.3 s, 42 m/s)
Here is what you start with:
 X = 300 ft or 91.44 m
 t = ?
 V_{o} = 0 (he started at rest)
 V_{a} = ?
 V = ?
 a = 9.8 m/s/s (he's going down)
We find a formula
X = V_{o}t + ^{1}/_{2}at^{2} seems like the best one
So we put in the numbers:
91.44 = (0)t + 1/_{2}(9.8 m/s^{2})t^{2}
91.44 m = 4.9t^{2}
18.661 = t^{2}
with the square root we find t = 4.3 seconds
Now moving on to impact velocity
Using the formula V_{f} = V_{o} + at
Plug it in, Plug it in aaannnnddd
V_{f} = 42.238 or 42 m/s
Go to: Problem Formulas Table of Contents
4.
Red Elk leaves the 10.0 m diving board with an
upward velocity and hits the water 1.9 seconds later. What was his initial upward velocity?
To what height above the diving board did he rise before going down?
With what speed did he hit the water?
(4.0 m/s, .84 m, 14.6 m/s)
Here is what you start with:
 X = 10.0 m
 t = 1.9 seconds
 V_{o} = ?
 V_{a} = ?
 V = ?
 a = 9.80 m/s/s (The acceleration of free fall on Earth)
We first have to find an appropriate formula.
Since weíre solving for V_{o} letís find one that solveís for it.
X = V_{o}t + ^{1}/_{2}at^{2} is quite appetizing letís see if it solves only for V_{o}
10m = V_{o}(1.9s)+ ^{1}/_{2}(9.8 m/s/s)(1.9)^{2}
those arithmetic skills yield V_{o} = 4.0468 m/s which with sig figs is 4.0 m/s
Now, since we know that at the highest point his V = 0, we can use that to sovle for X
With 2aX = V^{2}  V_{o}^{2}
and the numbers make it look like this:
2(9.8 m/s^{2})X = 0^{2}  (4.04 m/s)^{2}
again we use those algebra skills and get X = .84 m
But weíre not done yet, we still must find Red Elkís velocity on impact
Weíll use the equation V_{f}^{2} = V_{o}^{2} + 2aX
Since the V_{o} is from the highest point then the V_{o} = 0
Then we fill in the rest of the equation V_{f}^{2} = 0 + 2(9.8 m/s^{2})(10.84)
Itís important to remember that the X = 10.84 because thatís the highest point of the arc
Because the original X = 10 m and the distance higher is .84 m
Working it through and remembering that itís negative because heís going down:
V_{f} = 14.576 or 14.6 m/s
Cool, itís done!!
Go to: Problem Formulas Table of Contents
5.
A car will skid to a halt at a rate of 9.4 m/s/s.
If you measure skid marks that are 34 m long, with what speed
was the car going that made them?
(25 m/s)
Here is what you start with:
 X = 34 m
 t = ?
 V_{o} = ?
 V_{a} = ?
 V = 0 m/s (the car is stopped)
 a = 9.4 m/s/s
This is a simple acceleration equation:
Since we donít have any time given the obvious equation is
V_{f}^{2} = V_{o}^{2} + 2aX
Weíre trying to find the original velocity an the plugged in numbers:
0^{2} = V_{o}^{2} + 2(9.4 m/s/s)(34 m)
That handy dandy math and V_{o} = 25.2824 or significantly 25 m/s!
Go to: Problem Formulas Table of Contents
6.
A train can speed up at .15 m/s/s. In what minimum
distance can it attain a speed of 25 m/s starting from rest?
(2083 m)
Here is what you start with:
 X = ?
 t = ?
 V_{o} = 0 (it tells us that it starts from rest)
 V_{a} = ?
 V = 25 m/s
 a = .15 m/s/s
Hey, Iíve got an idea, letís find a formula to use
We seem to use V_{f}^{2} = V_{o}^{2} + 2aX alot, and it seems best.
Linda, something from the meat cabinet. Or perhaps the number cabinet.
25^{2} = 0^{2} + 2(.15 m/s/s)X
Break it down homeboy, and you get X = 2083.3333333333333333333 or 2083 m
Go to: Problem Formulas Table of Contents
7.
A drag racer can reach a speed of 53 m/s over
a distance of 120 m. What is its acceleration? Over what distance
can it reach a speed of 85 m/s?
(11.7 m/s/s, 309 m)
Here is what you start with:
 X = 120 m
 t = ?
 V_{o} = 0 m/s
 V_{a} = ?
 V = 53 m/s
 a = ?
I tell ya, I just canít get enough of that V_{f}^{2} = V_{o}^{2} + 2aX
With some semiredundant plugging in of numbers we have
(53 m/s)^{2} = 0^{2} + 2a(120 m)
Hey, uhhhh, letís try solving it through!!
Lo and behold, a = 11.7 m/s/s
What, thereís more, letís see what formula might be a good idea.
Since we donít know t weíll use the formula:
r_{2} = ^{1}/_{(mv2/2ke2) + (1/r1)}
Ha, Ha, Ha, thatís not until electric energy.
Instead we use good oleí V_{f}^{2} = V_{o}^{2} + 2aX
Those numbers plugged in, remembering that the V_{f} = 85 m/s gives us
(85 m/s)^{2} = 0^{2} + 2(11.7 m/s/s)X
And if we can still remember how to do algebra, than we find X = 308.76068 or 309 m
Go to: Problem Formulas Table of Contents
8.
A jetliner must reach a speed of 80 m/s to take off,
and can accelerate at 4.7 m/s/s. What is the minimum length of runway?
(681 m)
Here is what you start with:
 X = ?
 t = ?
 V_{o} = 0 (from rest)
 V_{a} = ?
 V = 80 m/s
 a = 4.7 m/s/s
Since we donít know the amount of time than we must again use
V_{f}^{2} = V_{o}^{2} + 2aX
Those numbers go in and make the lovely little equation
(80 m/s)^{2} = 0^{2} + 2(4.7 m/s/s)X
Work it on down and you are left with X = 680.8510638 or 681 m
Go to: Problem Formulas Table of Contents
9.
Theoretically, what would be the velocity of a steel marble dropped from an airplane
1000 m above the ground just as it hits the ground?
(140 m/s)
Here is what you start with:
 X = 1000 m
 t = ?
 V_{o} = 0 m/s
 V_{a} = ?
 V = ?
 a = 9.8 m/s/s
Use the formula V_{f}^{2} = V_{o}^{2} + 2aX
This formula seems best for most of these problems because they donít give you time
Anyway, plug the stuff in to have
V_{f}^{2} = 0^{2} + 2(9.8 m/s/s)(1000 m)
bust a move and get V_{f} = 140 or 140 m/s (because itís going down!)
Go to: Problem Formulas Table of Contents
10.
A rifle bullet leaves the muzzle of a .75 m long barrel going 450 m/s.
What is the acceleration of the bullet while it is in the barrel?
(135,000 m/s/s)
Here is what you start with:
 X = .75 m
 t = ?
 V_{o} = 0 m/s
 V_{a} = ?
 V = 450 m/s
 a = ?
Use V_{f}^{2} = V_{o}^{2} + 2aX
We then have (450 m/s)^{2} = (0 m/s)^{2} + 2a(.75 m)
202500 = 1.5a
a = 135000 m/s/s
Go to: Problem Formulas Table of Contents
Hey this page was not easy, but I feel priviledged to have been able to help you grasp
Physics. And I really needed the extra credit! Happy Solving!!