A truck covers 45.0 m in 8.80 s while smoothly slowing down to final speed of 3.00 m/s. Find Its Original Speed.

7.23 m/s

Explanation:

From Newton's equation of motion,

v = u + at ... Equation 1.

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

Also,

s = ut + 1/2at² ... Equation 2

Where s = distance.

Given: t = 8.8 s, s = 45.0 m.

Substitute into equation 2

Note: we find the value of a in terms of u

45 = u (8.8) + 1/2a (8.8) ²

45 = 8.8u+38.72a

38.72a = 45 - 8.8u

38.72a = (45-8.8u)

a = (45-8.8u) / 38.72

also, v = 3.00 m/s

Substituting into equation 1

3 = u + 8.8[ (45-8.8u) / 38.72) ]

3 = u + (45-8.8u) / 4.4

3*4.4 = 4.4u + 45 - 8.8u

13.2 - 45 = 4.4u - 8.8u

-31.8 = - 4.4u

u = - 31.8/-4.4

u = 7.23 m/s.

Hence the initial velocity = 7.23 m/s