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A car crosses 10m distance from point A to point B in 1 second, next 10m distance from point B to point C it crosses in 0.8s. Having distances and times is it possible to calculate acceleration and speed at points A,B,C?

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No, it is not possible since there are multiple ways in which a car could achieve these times using different speeds.

It could, for example stay at point A for 0.5 seconds not moving at all. Then travel at a speed of 20m/s for exactly 0.5 seconds passing point B at 1 second. From point B on it could travel at 12.5 m/s reaching point C after 1.8s as wished. The same can however be achieved by just not waiting at the beginning.The car could travel at 10m/s from point A to B and then at 12.5 from B to C. This does also satisfy the times that you stated but the car crosses the B point only half as fast.

That said as your question is tagged as quadradics it is possible to determine the speeds assuming constant acceleration. Under constant acceleration a the positions s after time t is given by:

s(t) = 0.5a*t^2 + v_0*t+s_0

We already now that s(0) = 0 and that s(1) = 10m and s(1.8) = 20m. plugging these into the formula we get two equations.

I.: s(1) = 10m = 0.5a*(1s)^2 + v_0*1s

II.: s(1.8) = 20m = 0.5*a*(1.8s)^2 + v_0 * 1.8s

These are two equations with two unknowns, namely v_0 and a. Solving this system of equations will yield the correct answers.

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  • $\begingroup$ Thanks, that was a great help. $\endgroup$ – Zijo Sep 16 '16 at 2:03
  • $\begingroup$ Sorry I was looking for the way how to do it but I did not find it. Now I looked again. $\endgroup$ – Zijo Sep 17 '16 at 1:23
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yes it is, you can make two equations s follow; let velocity of car at A is 'u' and acceleration be 'a',then From A to B: 10=ut+(at^2)/2 or putting t=1 you will get 10=u+a/2: Also for A to C: 20=ut+(at^2)/2 or putting t=1.8(1+0.8) you will get 20=1.8u+1.62a; you have two variables u and a and two equations,solve them and get your answer.

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