2
$\begingroup$

A

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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