Accelerated motion - mechanics

The delivery truck, with a total weight of 3.6 t, accelerates from 76km/h to 130km/h in the 0.286 km long way. How much was the force needed to achieve this acceleration?

Correct answer:

F =  5402.0979 N

Step-by-step explanation:

m=3.6 t kg=3.6 1000  kg=3600 kg s=0.286 km m=0.286 1000  m=286 m v1=76 km/h m/s=76:3.6  m/s=21.11111 m/s v2=130 km/h m/s=130:3.6  m/s=36.11111 m/s  s = v1 t + 21a t2 a =  tv2v1  s = v1 t + 21 tv2v1 t2 s = v1 t + 21(v2v1) t  t=v1+(v2v1)/2s=9190+(93259190)/2286=51551489.9961 s  a=tv2v1=9.996136.111121.11111.5006 m/s2  s1=v1 t+21 a t2=21.1111 9.9961+21 1.5006 9.99612=286 m s1=s  F=m a=3600 1.5006=5402.0979 N

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