Question #167966

A 2600 kg truck traveling to the north is slowed down uniformly from an initial velocity of 29 m/s by a 6350 N braking force acting opposite the truck's motion.

The truck's velocity after 1.6 seconds is   m/s.

The time is takes the truck to come to a complete stop is 


1
Expert's answer
2021-03-02T18:05:35-0500

(a)

FΔt=m(vfvi),F\Delta t=m(v_f-v_i),FΔt=mvfmvi,F\Delta t=mv_f-mv_i,vf=FΔt+mvim,v_f=\dfrac{F\Delta t+mv_i}{m},vf=6350 N1.6 s+2600 kg29 ms2600 kg=25.1 ms.v_f=\dfrac{-6350\ N\cdot1.6\ s+2600\ kg\cdot29\ \dfrac{m}{s}}{2600\ kg}=25.1\ \dfrac{m}{s}.

(b)

FΔt=m(vfvi)F\Delta t=m(v_f-v_i)Δt=2600 kg(029 ms)6350 N=11.9 s.\Delta t=\dfrac{2600\ kg\cdot(0-29\ \dfrac{m}{s})}{-6350\ N}=11.9\ s.

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