Suppose that the stopping distance D of a car after the breaches have been applied varies directly as the square of the speed s. A certain car travelling at 50mih can stop in 240 ft. What is the maximum speed it can be travelling if it needs to stop in 160ft?
D=ks2Model240ft×1mile5280ft=k(50mih)2Substitute the given and convert ft into miles122mi=k(2500mi2h2)Simplifyk=122mi2500mi2h2k=155000h2mi
Then if,
D=160ft,D=ks2160ft×1mile5280ft=155000h2mi(s2)Solve for s133mi=155000h2mi(s2)s2=5500033mi2h2Take the square roots=50√63mih
It shows that if the car needs to stop in 160ft. Its speed must be a maximum of 50√63mih
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