Kinematics problem with constant acceleration.

AI Thread Summary
A car with constant acceleration travels 56.1 meters in 5.44 seconds, reaching a speed of 14.2 m/s at the second point. To find the initial speed and acceleration, users suggest using kinematic equations, particularly focusing on the average speed and the relationship between distance, time, and acceleration. There is confusion about the correct application of these equations, with some users expressing frustration over their calculations. A proposed method involves solving for the initial speed using the average speed formula and then determining acceleration from there. The discussion emphasizes the need for two equations to solve for the unknowns effectively.
motoxxx
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Homework Statement


A car moving with a constant acceleration covered the distance between two points 56.1 m apart in 5.44s. Its speed as it passes the second point was 14.2m/s. (a) what was the speed at the first point? (b) What was the acceleration?. (c) At what prior distance from the first point was the car at rest?


Homework Equations



These are all I came up with. This one has me stumped.

vf = vi + a delta t ------ (a)

ax = v2x - v1x --------(b)
_________
t2 - t1

delta x = 1/2 (vf +vi ) delta t -------- (c)




The Attempt at a Solution



(a)vf = 56.1 + x(delta t)
=?

(b) assuming that acceleration is constant, then between the two known points would it be 10.31 m/s^2?

OR

v avg = 14.2 +0
_______
2
= 7.1 m/s

Therefore,

ax = vx * t
= 7.1 * 5.22
=39.22 m/s^2

(c) ------ No idea where to start.

Help!
 
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Can anyone point me in the right direction?
 
I can try to help you. Do you know any other kinematic equations? Of the equations you posted, b looks similar to a, just rearranged, and I don't know where the x came from, and c is totally foreign to me.

A couple of key points, velocity is changing and I don't think an average velocity does you any good.

I don't quite understand what you did. I've done the calculations and have what seems to be a correct answer (I put my answers back to test them). To begin with, you need vi and a, two unknowns, so you want to find two equations to use, then solve for those unknowns. Once that is done, you can do part c.

That's my suggestion, someone else may have a better idea.
 
I don't even know how to derive the equations for this type of question. Is there any way you can show an example of what you did to get me started?

thanks
 
They are:

\Delta x = v_1t + \frac{1}{2}at^2

vf = vi + at
vf^2 = vi^2 + 2a \Delta x

The second one is the same as your 'a' equation. You can see that you don't have quite enough information to complete any of them, but if you start filling stuff in, you should be able to get to a point where you have two equations with two unknowns. This will get you a and b of your problem. Then, to solve for c, for now I'm going to set you off on your own. If you run into more trouble I'll help.
 
Man, since you wrote that reply, I have been plugging in numbers every which way and I I am about ready to throw my textbook out the god damn window!

I have substituted equations for the unkown's left right and center and gotten nowhere.

I have used the equations you provided and have not been successful at all.

For the love of god, show me the answers!
 
motoxxx said:
Man, since you wrote that reply, I have been plugging in numbers every which way and I I am about ready to throw my textbook out the god damn window!

I have substituted equations for the unkown's left right and center and gotten nowhere.

I have used the equations you provided and have not been successful at all.

For the love of god, show me the answers!

for part a)

d = [(v1+v2)/2]*t

solve for v1.

there are different ways to do the problems, but this one is probably the most direct way.
 

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