Question

A truck covers 40.0 m in 9.00 s while uniformly slowing down to a final velocity of 2.20 m/s.(a) Find the truck's original speed. m/s(b) Find its acceleration. m/s2

Answer 1

Given:

The distance covered by truck: d = 40.0 m

The time taken to cover the distance is: t = 9.00 s

The final velocity of the truck is: v2 = 2.20 m/s

To find:

a) the speed of the truck.

b) the acceleration

Step-by-step explanation:

a)

The speed of the truck before it slows down can be calculated as:

`d=(1)/(2)(v_2+v_1)t`

Substituting the values in the above equation, we get

`\begin{gathered} 40=(1)/(2)(2.20+v_1)*9 \\ \\ (40*2)/(9)-2.20=v_1 \\ \\ v_1=6.69\text{ m/s} \end{gathered}`

b)

The truck is initially moving at a speed of 6.69 m/s. It then slows down to the final velocity of 2.20 m/s. The acceleration of the truck can be determined as:

`d=v_1t+(1)/(2)at^2`

Substituting the values in the above equation, we get:

`\begin{gathered} 40=6.69*9+(1)/(2)* a*9^2 \\ \\ 40=60.21+40.5a \\ \\ a=(40-60.21)/(40.5) \\ \\ a=-0.499 \\ \\ a\approx-0.5\text{ m/s}^2 \end{gathered}`

Final answer:

a) The original speed of the truck is 6.69 m/s.

b) The acceleration of the truck is - 0.5 m/s^2.