# Concerning motion

1. Mar 6, 2005

### Kamataat

Hi, I was given the following question:

Q: A body starts moving with an acceleration $a=const$. After a time $t$ it has covered the distance $$\Delta x[/itex]. What is its speed at time $t$? Answer: $$\frac{2\Delta x}{t}$$ Can somebody say how to arrive at this answer (both mathematically and intuitionally)? I know that $\Delta x/t$$ would be the speed w/o acceleration, so the "2" must come from acceleration, but where does it come from? Can the answer be obtained from the formula [itex]v=v_0+at^2$?

- Kamataat

2. Mar 6, 2005

### scholzie

$$x-x_0=\frac{1}{2}(v_0+v_f)\Delta{t},\ \ v_0=0 m/s$$

3. Mar 6, 2005

Thank you!

- Kamataat

4. Mar 6, 2005

### HallsofIvy

Staff Emeritus
Another way: for CONSTANT acceleration, the "average" speed is just the average of the starting speed and ending speed: $\frac{v_i+v_f}{2}$. Since the starting speed is 0, if the ending speed is v, this is $\frac{v}{2}$ so the distance covered in time $\Delta t$ would be $\Delta x =\frac{v}{2}\Delta t}$ so $v= 2\frac{\Delta x}{\Delta t}$.

CAUTION: This is only true for CONSTANT acceleration!

Last edited: Mar 6, 2005
5. Mar 7, 2005

### scholzie

Halls, that's what my equation says :) $1/2(v_0 + v_f) =$ average velocity

6. Mar 7, 2005

### Kamataat

yup, i get it now. tnx