Question

A truck covers 40.0 m in 9.15 s while uniformly slowing down to a final velocity of 3.00 m/s.

(a) Find the truck's original speed.
m/s

(b) Find its acceleration.
m/s²

(Can you please show your work, thank you)

Answer 1

Given : x = 40 m ; t = 9.15 s ; v = 3 m/s

Goal : v₀ =? ; a =?

Solution :

Applying equation of motion as :

v = v₀ + at

3 = v₀ + 9.15 a ...........(1)

And

x = v₀t + (1/2)at²

40 = 9.15v₀ +41.86 a ..........(2)

From equation (1) and (2) we have :

40 = 9.15(3-9.15a)+ 41.86 a  

12.55 = -41.86 a

a = -0.3 m/s²

putting value of a in equation (1) we get

v₀ = 3- 9.15(-0.3) = 5.745 m/s