# Acceleration problem.

1. Jan 24, 2014

### patelneel1994

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)

2. Relevant equations

Vf = Vo + at

3. 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

Last edited: Jan 24, 2014
2. Jan 24, 2014

### Apothem

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.

Last edited: Jan 24, 2014
3. Jan 24, 2014

### BvU

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. Jan 24, 2014

### BvU

I missed your response because it came to me as a private message:
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. Jan 24, 2014

### BvU

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!

6. Jan 25, 2014

### patelneel1994

Thank you BvU.
Both of your help is appreciated.