# Finding when an object overtakes another object using a speed-time graph 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.

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

• 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.