Answer to Question #342295 in Physics for Muyunda

Question

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? [

Answer 1

(a)

$$v=v_0+at$$

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

(b)

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

(c)

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