A truck covers 40 meters in a 8.50 seconds while smoothly slowing down to a final velocity of 2.80 meters per second. Perform the following tasks:

a) find the trucks original speed

b) find its acceleration

s = (v-u) / 2 * t, where v is the final velocity, u is the initial velocity, s is the displacement and t is the time.

40 = (v-2.8) / 2 * 8.5

4.71 = v-2.8/2

9.41 = v - 2.8

v = 12.2 m/s

v^2 = u ^2 + 2as where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

12.2^2 = 2.8^2 + 2a (40)

80a = 12.2^2 - 2.8^2

80a = 148.84 - 7.84

80a = 141

a = 1.7625m/s^2