Acceleration problem.

  • #1
1. The problem statement,
In 2.0 s, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 50 m. The velocity at the end of this time interval is 10 m/s. What is the acceleration of the particle?

all variables and given/known data

t = 2.0 s, constant acceleration , disttance = 50 - 10 = 40 m, Vf = 10 m/s(I'm not sure if this would be cosidered as fenal velocity or not)


Homework Equations



Vf = Vo + at

The Attempt at a Solution


I don't know the Vo.
I've tried doing it using distance( d = vi.t + (1/2 . a.(t^2)) formula didn't work.
Thank you
 
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Answers and Replies

  • #2
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The particle did not move a distance of 10m, look at what you did, I think you just made a silly mistake. You are right when you say the final velocity is 10m/s.

If I say another SUVAT equation:

s=[(Vf+Vo)/2]*t

where s is the distance

Do you think you would be able to calculate Vo? From this you can just use the equation you stated to find the acceleration.
 
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  • #3
BvU
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You are doing well!
You have two equations:
one: x(t) = x0 + v0 * t + 1/2 * a * t^2 with x0 = x(0) = 10 m, given is x(2) = 50
two: v(t) = v0 + a * t and given is v(2) = 10 m/s

As you say, v0 is unknown. And a is another unknown. Two equations with two unknowns generally can be solved !
 
  • #4
BvU
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I missed your response because it came to me as a private message:
patelneel1994 said:
Can you eleborate those two formulas please!
what is Xo?

If you could, give me some useful formulas to solve these kinds of problems and good stratagy.
THank you
Well, the formulas are there in #3 and the strategy is to interpret them correctly and then solve them.

That probably isn't helping you enough, so let me try it a bit more elaborate:

Something is described that takes two seconds. Let's say from t=0 s to t=2 s. This sets the time scale, just as you have done already.

At t=0 s, (t0) the position on the x-axis is x(0) = x0 = 10 m This is where we start when we start.
We don't know the speed at this point, so we call it v0. v0 = v(0) .

At t=2 s, the position on the x-axis is x(2) = 50 m
We know the speed at this point and time, v(2) = 10 m/s .

We have expressions for linear motion with constant acceleration:
1) for the position as a function of time
x(t) = x0 + v0 * t + 1/2 * a * t^2
2) for the speed in the x-direction as a function of time
v(t) = v0 + a * t

I have used symbols x0 and v0 for the known x0 and the (still) unknown v0

We fill in what we know at time t = 2 s:
x(2) = 50 m = 10 m + v0 * 2 s + 1/2 * a * (2 s)^2
v(2) = 10 m/s = v0 + a * 2 s

This is two equations with two unknowns. The exercise asks for the acceleration a, so we eliminate v0 from the first equation by using the second equation: The latter gives us
v0 = 10 m/s - a * 2 s
which gives us one equation with one unknown:
50 m = 10 m + (10 m/s - a * 2 s) * 2 s + 1/2 * a * (2 s)^2

writing it out and bringing all knowns to one side gives us a !
 
  • #5
BvU
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I notice Doc Al has helped you nicely with the balcony exercise. I'm going to sleep now. You are doing well, keep it up!
 
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  • #6
Thank you BvU.
Both of your help is appreciated.
 

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