# Motion question that sounds too simple?

1. Aug 8, 2010

### Pronitron

1. The problem statement, all variables and given/known data
A car accelerates steadily from 0 ms^-1 to 20 ms^-1 in a distance d and time t. Another car takes a time 2t to accelerate steadily from stationary to the same final velocity. What distance does the car travel during the new acceleration?

2. Relevant equations
So Speed x Time = Distance
And then there are the non-linear acceleration ones but I'm not sure if they are needed

3. The attempt at a solution
Okay so intuitively I think that because it's taken double the time it has travelled double the distance, because if the time is 2t then to get the same speed we need 2d, but I don't think the question says that both cars travel at constant acceleration (i.e the same) so it could be that one accelerates at a greater rate than the other or something? What do you think of the wording?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 8, 2010

### Pengwuino

The key to the problem is "steadily", meaning a constant acceleration. If two objects accelerate to the same velocity, but one takes twice the time, they will have different (but still constant) accelerations. Consider from rest, with no initial velocity, a object travels a distance $$d = \frac{at^2}{2}$$. What does the situation tell you about how 'a' and 't' change going from one situation to the next?