A car traveling at 48mph overtakes a cyclist who, riding at 12 mph, haf at 3-hr head start. How far from the starting point does the car overtakes the cyclist?

Question

anonymous · Answer

Let $t$ be in hours and when the cyclist begins his journey.
Position of the car: $x_\text{car}=v_\text{car}(t-3)$
Position of the cyclist: $x_\text{cyclist}=v_\text{cyclist}t$
Find out when we have $x_\text{car}=x_\text{cyclist}$, and then use that to answer the question.

anonymous · Answer

Do I need to add both speed 48 and 12?

anonymous · Answer

No.  The velocities are independent.

anonymous · Answer

distance between them=36 miles
and relative speed = 48-12=36

so time taken =36/36=1hr

anonymous · Answer

I don't understand it. Can u please explain to me. I really appreciate it

anonymous · Answer

so what happen to three hours?

anonymous · Answer

see the the cyclist started riding at 12 miles per hour before the car and the car started after 3 hours...so in 3 hours distance travelled by cyclist=12*3=36miles

anonymous · Answer

So u mean in one hour the car overtake the cyclists?

anonymous · Answer

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