Question

A truck covers 35.0 m in 8.70 s while smoothly slowing down to final speed of 2.70 m/s. (a) find its original speed. (b) find its acceleration. (indicate the direction of the acceleration with the sign of your answer.)

Answer 1

35/ 8.70 = 4.022

4.022x 2 = 8.044

8.044 - 2.7 = u = 5.34 m/s

(u + v) / 2 x t = s = 35

(5.34 + 2.7) / 2 x 8.9 = 35

(b) Find its acceleration. (Indicate the direction of the acceleration with the sign of your answer.)

Answer must be in m/s^2

v^2 = u^2 + 2as

v^2 = 2.7^2 = 7.29

u^2 = 8.41^2 = 70.7281

s = 35m

70.7281 – 7.29 = 2as

63.9681 = 2as

63.4381/(2 x s) = a = 63.4381/(2 x 35) = 0.9062 <---- Reverse into minus

v^2 = u^2 + 2as

6.76 = 70.7281 + ( 2 x (-0.9062) x 49)

Acc = -18.08m/s^2