The driver of an empty speeding truck slams on the brakesand skids to a stop in

Question

The driver of an empty speeding truck slams on the brakesand skids to a stop in adistance Dm = 254.84 m. Thetruck's initial velocity is v0 = 50m/s and its total mass is m = 3137.8kg. The coefficient of kinetic friction between the tiresand the road is mk = 0.5

If the truck were carrying an amount of cargothat doubled its total mass, what would itsstopping distance have been? (Assume the road is straight andhorizontal).

D2m = m

I tried using the equation fk=uk*N and thus, -ma=uk*mg and triedto solve for acceleration, but I'm lost as to what equation I woulduse next.

Explanation / Answer

now mass=3137.8*3=9413.4 Kg (since cargo weigh 2*3137.8 and thetruck weigh 3137.8) g=accleration due to gravity=9.8 frictional force =*9413.4*g=0.5*9314.4*9.8=46125.66 negative accleration caused by the frictional force=4.9m/s2 let the distance needed to stop the truck=s u=50 m/s v2=u2-2*a*s 0=502-2*4.9*s s=255.10 m (ans) or, aprox 255 m D2m=255.1 m

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