Question #349295

03*. A 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? [9.91s]


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


(c) How far does it travel during the third second after the brakes are applied? [16.8 m]

1
Expert's answer
2022-06-09T09:59:22-0400

Given:

v0=22.5m/sv_0=22.5\:\rm m/s

a=2.27m/s2a=-2.27\:\rm m/s^2


(a)

vf=v0+at=0v_f=v_0+at=0

t=v0a=22.52.27=9.91st=-\frac{v_0}{a}=-\frac{22.5}{-2.27}=9.91\:\rm s

(b)

d=v0t+at2/2=22.59.91+(2.27)9.912/2=112md=v_0t+at^2/2\\ =22.5*9.91+(-2.27)*9.91^2/2=112\:\rm m

(c)

Δd=d(3)d(2)=(22.53+(2.27)32/2)(22.52+(2.27)22/2)=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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