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:
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: 
    

  1. 
  
    V = Vo + at

    

  2. 
  
    Va = (Vo + V)/2

    

  3. 
  
    X = Vat

    

  4. 
  
    X = Vot + 1/2at2

    

  5. 
  
    2aX = V2 - Vo2,  Which I like to have setup as:  Vf2 = Vo2 + 2aX


    




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General Problem Solving Strategy:





    

  1. Read the problem.

  2. Go through the problem and figure out what is given or implied
    
Make a list, and identify the quantities you know.

  3. Find any formula that will allow you to calculate 
    
anything that you don't know, and apply it.

  4. Add what you just found in the last step to your list of knowns.

  5. Check to see if you have found the answer.  If not, repeat the 
    
previous two steps until you are done.




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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)



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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)


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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:

Well let's find an equation that solves for X

Of the many that do fit, 2aX = V2 - Vo2 is best because we know all the variables exept for X.

Plug in those cute numbers:

2(-9.8 m/s2)X = (0)2 - (54 m/s)2

X  =   -2916 m/s
      -------------
        -19.6 m/s2

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 = Vot + 1/2at2 looks appealing so let's try it.



We then plug in the numbers: 


149 = 0t + 1/2(-9.8 m/s2)t2

Finishing it through:

149  = -4.9t2

Leaving us with  t2  = 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:

Let's plug this in to the formula that we know all the variables for, except for t

V = Vo + at,  seems the most appropriate

So, -24 m/s = (0) + (-9.8m/s2)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 = Vat

Va  = -12 m/s because Va = (Vo + 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:

We find a formula

X = Vot + 1/2at2 seems like the best one

So we put in the numbers:

91.44 = (0)t + 1/2(9.8 m/s2)t2

91.44 m = 4.9t2

18.661 = t2

with the square root we find t = 4.3 seconds

Now moving on to impact velocity

Using the formula Vf = Vo + at

Plug it in, Plug it in  aaannnnddd

Vf = 42.238 or 42 m/s

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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:


We first have to find an appropriate formula.  


Since weíre solving for Vo letís find one that solveís for it.


X = Vot + 1/2at2  is quite appetizing letís see if it solves only for Vo 


10m  = Vo(1.9s)+ 1/2(-9.8 m/s/s)(1.9)2

those arithmetic skills yield Vo  = 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 = V2 - Vo2

and the numbers make it look like this:

2(-9.8 m/s2)X  = 02 - (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 Vf2  = Vo2 + 2aX

Since the Vo is from the highest point then the Vo = 0

Then we fill in the rest of the equation Vf2 = 0 + 2(-9.8 m/s2)(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:
Vf = -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:


This is a simple acceleration equation:

Since we donít have any time given the obvious equation is

Vf2 = Vo2 + 2aX

Weíre trying to find the original velocity an the plugged in numbers:

02 = Vo2 + 2(-9.4 m/s/s)(34 m)

That handy dandy math and Vo = 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:


Hey, Iíve got an idea, letís find a formula to use

We seem to use Vf2 = Vo2 + 2aX alot, and it seems best.


Linda, something from the meat cabinet.  Or perhaps the number cabinet.

252 = 02 + 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:


I tell ya, I just canít get enough of that Vf2 = Vo2 + 2aX

With some semi-redundant plugging in of numbers we have

(53 m/s)2 = 02 + 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:

r2 = 1/(mv2/2ke2) + (1/r1)

Ha, Ha, Ha, thatís not until electric energy.

Instead we use good oleí Vf2 = Vo2 + 2aX

Those numbers plugged in, remembering that the Vf = 85 m/s gives us

(85 m/s)2 = 02 + 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:


Since we donít know the amount of time than we must again use

Vf2 = Vo2 + 2aX

Those numbers go in and make the lovely little equation

(80 m/s)2 = 02 + 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:


Use the formula Vf2 = Vo2 + 2aX

This formula seems best for most of these problems because they donít give you time

Anyway, plug the stuff in to have

Vf2 = 02 + 2(9.8 m/s/s)(1000 m)

bust a move and get Vf = 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:

Use  Vf2 = Vo2 + 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!!