Calculating acceleration when only have time and distance

  • Thread starter Thread starter artworkmonkey
  • Start date Start date
  • Tags Tags
    Acceleration Time
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 4K views
artworkmonkey
Messages
12
Reaction score
1

Homework Statement


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?

Homework Equations


None given

The Attempt at a Solution


Using only time and distance traveled I can only think how to calculate the average velocity: Vave = distance traveled / 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
 
Physics news on Phys.org
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?
 
  • Like
Likes   Reactions: 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 :)
 

Attachments

  • Time vs Velocity.png
    Time vs Velocity.png
    17 KB · Views: 686
Very close indeed. However: ##distance = \frac{1}{2} Acceleration \cdot time^2##. So fix that up and you should be good.
 
  • Like
Likes   Reactions: 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!