A problem says :

A car starts from rest and undergoes a constant acceleration. It travels 5 m in the t interval 0 s to 1 s, find the displacement of the car during the time interval from 1 s to 2 s?

the answer is 15m and I wonder why it is!

Because the acceleration in the interval from 0 to 1 will be 5 m/s^2 ..and from 1 to 2 will be 10 m/s^2

Therefore the displacement will be 10m not 15. What do you think?


closed as off-topic by Danu, Bernhard, Kyle Oman, ACuriousMind, Emilio Pisanty Sep 8 '14 at 21:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Danu, Kyle Oman, Emilio Pisanty
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ The acceleration will not be 5 m/s$^2$. Use $s=ut+\frac{1}{2}at^2$ to find $a$ where you know that $u=0$, $s=5$ and $t=1$. $\endgroup$ – lemon Sep 8 '14 at 20:12
  • 2
    $\begingroup$ Think! Your first sentence says "constant acceleration", but later on you say that the acceleration has changed. You can't have it both ways! $\endgroup$ – garyp Sep 8 '14 at 20:15
  • $\begingroup$ @garyp Oh ! i mean velocity not acceleration but i dont know why i wrote acceleration . Anyway.. first the velocity is 5 m/s .. then in the 2nd interval subsequently it will be 10 m/s .. it looks correct ! $\endgroup$ – Maher Sep 8 '14 at 20:54
  • $\begingroup$ I know it is wrong ^.. but how do you explain that a=10 practically ( without using formulas) !!!?? $\endgroup$ – Maher Sep 8 '14 at 20:56
  • $\begingroup$ This might be a duplicate: What is the intuition behind $gt^2/2$ equalling the distance something falls after $t$ seconds? $\endgroup$ – Bernhard Sep 8 '14 at 21:00

Displacement is

$$x = \frac12 a t^2$$

Without solving for acceleration, the ratio of displacements for two times is

$$\frac{x_1}{x_2} = \frac{t_1^2}{t_2^2}$$

So if $x_1 = 5$, then it follows from the above that $x_2=20$ and the difference (the distance covered in the second time interval) is $15 m$.


If it undergoes constant acceleration $a$ from a stop, then at time $t$ seconds it will have traveled ${a t^2}/2$ distance.

So substitute: if ${a 1^2}/2$ equals $5$ meters, then $a$ equals $2 * 5$ equals $10m/s^2$.

You take it from there.

  • $\begingroup$ ! But when i look at it as .. the velocity changed from 0m/s at rest to 5m/s at t=1 ( because it travelled a distance of 5 m in 1 s period ) therefore we can say that there is acceleration = 5? How to prove that thia idea is wrong? It is stuck in my mind :( $\endgroup$ – Maher Sep 8 '14 at 21:00
  • $\begingroup$ This** instead of "thia" $\endgroup$ – Maher Sep 8 '14 at 21:03
  • 1
    $\begingroup$ @Maher: Yes, its average speed was 5m/s, but it did not travel the whole distance at that speed. At the start of the second, its speed was zero, right? At the end of the second, its speed has to be larger than 5m/s, right? because it it was 0 at the start, and 5 at the end, its average would have to be less than 5, right? The average is 5, but the speed changes from 0 at the start to something more than 5 at the end, 10 in fact. Keep in mind that acceleration is the amount that speed changes every second. That's why it is meters per second per second. $\endgroup$ – Mike Dunlavey Sep 8 '14 at 21:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.