# Help with deriving time, velocity, displacement

1. Jan 5, 2015

### RachaelA

1. The problem statement, all variables and given/known data
A dog accelerates from 2.0 m/sec to 5.0 m/sec while covering a distance of 14.0 m. How long did this take

2. Relevant equations
d=(Vi+Vf/2)t

3. The attempt at a solution
2(-Vi-Vf)d=t

2. Jan 5, 2015

### Cake

Your algebra is off. You shouldn't be subtracting the sum of the velocities. In the equation (Vi+Vf) is being multiplied into time "t." So what should you actually do to move it to the other side?

3. Jan 5, 2015

### RachaelA

I think I should divide it?

4. Jan 6, 2015

### Maged Saeed

Is this the formula to find the time ??

You have to put the acceleration in consideration ,

Try to Find the acceleration first [I think it is constant] then the time from the equations of motion with constant acceleration.

:)

Last edited: Jan 6, 2015
5. Jan 6, 2015

### vktsn0303

I don't understand why you used
2(-Vi-Vf)d=t

t is not velocity/distance if there is acceleration.

I am assuming the acceleration to be a constant here. Use the laws of motion to find the acceleration. Once acceleration is determined it is easy to find how long the dog took to reach a final velocity.

6. Jan 6, 2015

### vktsn0303

Sorry. A small correction in my previous reply.
t is not velocity*distance.

7. Jan 6, 2015

### RachaelA

This is the answer according to my teacher

t=d/(Vf+Vi/2)

8. Jan 6, 2015

### haruspex

Seems like RachaelA has been taught to use the average velocity, being merely the average of initial and final velocities in the case of uniform acceleration. So no need to calculate the acceleration as such.
Please use parentheses correctly. What you have written means $d = \left(V_i+\frac{V_f}{2}\right)t$, but I hope you mean $d = \left(\frac{V_i+V_f}{2}\right)t$, which you could have written as d=((Vi+Vf)/2)t.
Yes.
Again, I hope you mean $t = \frac d{\left(\frac{V_i+V_f}{2}\right)}$ (which can be simplified a little, but no matter).
If so, dividing would have produced that.