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 result:

F =  5402.0979 N

Solution:

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+12at2 a=v2v1t  s=v1 t+12v2v1tt2 s=v1 t+12(v2v1)t  t=sv1+(v2v1)/2=28621.1111+(36.111121.1111)/2=51485159.9961 s  a=v2v1t=36.111121.11119.99611.5006 m/s2  s1=v1 t+12 a t2=21.1111 9.9961+12 1.5006 9.99612=286 m s1=s  F=m a=3600 1.5006=5402.0979 N



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