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Charles and Smith left a place abc at the same time and stated moving in the same direction.Charles speed was $15$ Kmph and Smith's speed was $12$ kmph.Half an hour later Anthony left the place and and travelled in the same direction.Sometime later he overtakes Smith and $90$ minutes further on he overtakes Charles.What is Anthony's speed?

Thanks for help!

haqnatural
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Tapadia Lion
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  • Is it a typo between "Kpmh" and "mph" ? Do you mean kilometer or mile ? – Zubzub Jul 28 '16 at 12:03
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    Welcome to MSE ! This site requires that you show the effort you have made. Knowing where you are stuck also helps us frame more suitable answers. – true blue anil Jul 28 '16 at 12:17

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The position of the people is the following. (Assuming Charles and Smith leave at $t=0$, and we use $km$ and $hour$ unit) $$ C(t) = 15t \\ S(t) = 12t \\ A(t) = v(t-0.5) $$ At some point they meet : $$ A(t) = S(t) \implies v(t-0.5) = 12t \\ A(t+1.5) = C(t+1.5) \implies v(t+1) = 15(t+1.5) $$ Solving these equations give $t = 1.5,\ v = 18$.

Zubzub
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    The first equation seems right but the second equation is flawed.You did not give an account for the distance travelled by Charles in that sometime when Anthony overcomes Smith. – Tapadia Lion Jul 28 '16 at 12:18
  • You are right ! Also $v=13$ is actually impossible since Anthony cannot overtake Charles... I edited ! – Zubzub Jul 28 '16 at 12:20