# Calculating acceleration when only have time and distance

1. Nov 30, 2015

### artworkmonkey

1. The problem statement, all variables and given/known data
A car travels 400m in 60s. The first 10s it accelerates from stationary. The last 10 seconds it decelerates back to stationary. For the middle 40s it has a constant velocity. What is the acceleration?

2. Relevant equations
None given

3. The attempt at a solution
Using only time and distance travelled I can only think how to calculate the average velocity: Vave = distance travelled / time = 6.67 m/s. I don't so much need an answer, more some guidance of where to begin. I'm pretty stuck. Thanks

2. Nov 30, 2015

### Staff: Mentor

Maybe start graphically? You might find some inspiration. Make a sketch of velocity versus time. You don't know the cruising speed yet, so leave that as a variable. Now, what do you know about the area under a velocity vs time graph?

3. Nov 30, 2015

### artworkmonkey

The area under the graph should equal the displacement, so the total area should equal 400m. I have drawn a graph (attached). When a put the acceleration and deceleration together the area is exactly one 5th of the time spent cruising: therefore, the amount of time spent accelerating and decelerating is one 5th of 400m = 80 meters. Divide this by 2 and the car moved 40 meters in 10 seconds.
Acceleration = distance/time^2 = 40m/10^2 = 0.4m/s

Does this sound right? Making a sketch was good advise because it is starting to make sense to me. Just hope I have taken a wrong turn somewhere.
Thanks :)

#### Attached Files:

• ###### Time vs Velocity.png
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27.3 KB
Views:
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4. Nov 30, 2015

### Staff: Mentor

Very close indeed. However: $distance = \frac{1}{2} Acceleration \cdot time^2$. So fix that up and you should be good.

5. Nov 30, 2015

### artworkmonkey

Yay! Thank you for your help. I've been staring at this question since yesterday, and actually drawing it was brilliant advice. Thank you!