Question #342295

truck travelling at 22.5 m/s decelerates at 2.27 m/s2



. (a) How much time does it take for the truck to stop? [9.91 ]s



(b) How far does it travel while stopping? [112m]



(c) How far does it travel during the third second after the brakes are applied? [

1
Expert's answer
2022-05-18T08:32:02-0400

(a)

v=v0+atv=v_0+at

t=vv0a=022.52.27=9.91st=\frac{v-v_0}{a}=\frac{0-22.5}{-2.27}=9.91\:\rm s

(b)

d=v2v022a=0222.522(2.27)=112md=\frac{v^2-v_0^2}{2a}=\frac{0^2-22.5^2}{2*(-2.27)}=112\:\rm m

(c)

Δd=d3d2=(22.532.2732/2)(22.522.2722/2)=16.8m\Delta d=d_3-d_2\\ =(22.5*3-2.27*3^2/2)\\ -(22.5*2-2.27*2^2/2)\\ =16.8\:\rm m


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