Answer to Question #349295 in Physics for Austine

Question

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]

Answer 1

Given:

$v_0=22.5\:\rm m/s$

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

(a)

$$v_f=v_0+at=0$$

$$t=-\frac{v_0}{a}=-\frac{22.5}{-2.27}=9.91\:\rm s$$

(b)

$$d=v_0t+at^2/2\\ =22.5*9.91+(-2.27)*9.91^2/2=112\:\rm m$$

(c)

$$\Delta d=d(3)-d(2)\\ =(22.5*3+(-2.27)*3^2/2)\\ -(22.5*2+(-2.27)*2^2/2)=16.8\:\rm m$$