The photo shows the speed-time graph of objects A and B for the first 20 seconds of their journey. Given that both objects start travelling at the same time on the same route, calculate the time taken by object B to overtake object A.

My workings: For overtaking to take place the distance travelled by A must be equal to the distance travelled by B.

Distance travelled by object A after 3 seconds = 3 m

Distance travelled by object B after 3 seconds = $\frac12×3×0.72=$ 1.08 m

Therefore overtaking happens some time after 3 seconds. After this I'm not sure how to find the time taken by B to overtake A.

  • $\begingroup$ Why did you take 3seconds? A overtakes B at $t=0$ $\endgroup$ – N.S.JOHN Oct 13 '16 at 8:37
  • $\begingroup$ But I'm trying to find when does B overtakes A again @N.S.JOHN $\endgroup$ – user307640 Oct 13 '16 at 8:41
  • $\begingroup$ See my ansswer. $\endgroup$ – N.S.JOHN Oct 13 '16 at 8:43

First of all, calculate how much distance the objects have travelled after 5 seconds, which is when both objects start travelling at constant speed.

  • A has travelled $5\cdot1=5$ metres
  • B has travelled $\frac12\cdot5\cdot1.2=3$ metres

B has not overtaken A by 5 seconds, so the overtaking must be at $5+x$ seconds where $x\ge0$. In $5+x$ seconds:

  • A has travelled $5+x$ metres
  • B has travelled $3+1.2x$ metres

Equating these two expressions gives $x+5=1.2x+3$, which rearranges to $0.2x=2$ or $x=10$. Therefore overtaking occurs at $10+5=15$ seconds.


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