# Answer to Question #15620 in Mechanics | Relativity for gail

Question #15620

a honda moving at 100 km/h is 80km behind a ford moving in the same direction at 60 km/h. how far does the honda travel before it catches up with the ford?

Expert's answer

Let's denote the distance traveled by honda as H, the distance traveled by ford as F and the time before meeting by T. Then

H = 100[km/h]*T

F =& 60[km/h]*T + 80[km]

H = F, therefore

100T = 60T + 80

40T = 80

T = 2[h]

So, honda will catch up the ford after two hours and it will travel 100[km/h]*2[h] = 200 kilometres.

H = 100[km/h]*T

F =& 60[km/h]*T + 80[km]

H = F, therefore

100T = 60T + 80

40T = 80

T = 2[h]

So, honda will catch up the ford after two hours and it will travel 100[km/h]*2[h] = 200 kilometres.

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