Question

A truck covers 40.0 m in 9.35 s while uniformly slowing down to a final velocity of 1.25 m/s.(a) Find the truck's original speed.____ m/s(b) Find its acceleration.____ m/s2

Answer 1

Given:

The distance covered by the truck is: x = 40 m

The time taken to cover the distance of 40 m is: t = 9.35 s

The final velocity of the truck is: v = 1.25 m/s

To find:

a) Truck's original velocity (initial velocity)

b) The acceleration of the truck

Explanation:

a)

The initial velocity "u" of the truck can be determined by using the following equation:

$\begin{gathered} x=\frac{v+u}{2}\times t \\ \\ \frac{2x}{t}=v+u \\ \\ u=\frac{2x}{t}-v \end{gathered}$

Substituting the values in the above equation, we get:

$\begin{gathered} u=\frac{2\text{ }\times40\text{ m}}{9.35\text{ s}}-1.25\text{ m/s} \\ \\ u=8.55\text{ m/s}-1.25\text{ m/s} \\ \\ u=7.3\text{ m/s} \end{gathered}$

The original speed of the truck is 7.3 m/s.

b)

The acceleration "a" of the truck can be calculated as:

$\begin{gathered} a=\frac{v-u}{t} \\ \\ a=\frac{1.25\text{ m/s}-7.3\text{ m/s}}{9.35\text{ s}} \\ \\ a=-\frac{6.05\text{ m/s}}{9.35\text{ s}} \\ \\ a=-0.65\text{ m/s}^2 \end{gathered}$

The acceleration of the truck is -0.65 m/s^2. The negative sign in the acceleration indicates that the truck is slowing down.

Final answer:

a) The original speed of the truck is 7.3 m/s.

b) The acceleration of the truck is -6.5 m/s^2.