# How do I calculate distance from acceleration and velocity?

1. Feb 7, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"An electric vehicle starts from rest and accelerates at a rate of $2.0 \frac{m}{s^2}$ in a straight line until it reaches a speed of $20 \frac{m}{s}$. The vehicle then slows at a constant rate of $1.0 \frac{m}{s^2}$ until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?"

2. Relevant equations
$v = v_0+at$
$x-x_0=v_0t+\frac{1}{2}at^2$

3. The attempt at a solution
(a)
$20\frac{m}{s} = 2\frac{m}{s^2}(t_1)$
$t_1=10s$
$0\frac{m}{s}=20\frac{m}{s}-1\frac{m}{s^2}(t_2)$
$-1\frac{m}{s^2}t_2=-20\frac{m}{s}$
$t_2=20s$
$t=t_1+t_2=10s+20s$

(b)
$x_1-0m=(0\frac{m}{s})(10s)+\frac{1}{2}(2\frac{m}{s^2})(10s)^2$
$x_1=100m$
$x_2-100m=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2$
$x_2=100m+400m-200m=300m$
$x=x_1+x_2=100m+300m ≠ 300m$

2. Feb 7, 2016

### ehild

x_2 is the position, not the displacement during the second stage of motion.

3. Feb 7, 2016

### akshay86

you have mistaken in attempting question b
only x2-100 gives the displacement in the time of disceleration and x2 is the total displacement and you calculated it has 300m(correct)

4. Feb 7, 2016

### PeroK

With a problem like this (two different acclerations) it can be useful to draw a graph of the motion: velocity against time. The displacement is then the area under the graph. It helps you visualise the motion and can prevent you from plugging the wrong numbers into your equations.

5. Feb 7, 2016

### Eclair_de_XII

So you're saying that x2 = x1 + x3, where x3 is some arbitrary distance between the end of x1 and x2? Then that would mean...

$x_2-100m=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2 = x_1+x_3-x_1 =x_3$
$x_3=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2=400m+(-200m)=200m$
$x_2=x_3+x_1=200m+100m=300m$

Thanks, guys.