Question

A truck covers 40.0 m in 8.55 s while uniformly slowing down to a final velocity of 2.45 m/s.(a) Find the truck's original speed. ________m/s(b) Find its acceleration. _______m/s2

Answer 1

Answer:

(a) 6.91 m/s

(b) -0.52 m/s²

Explanation:

Part (a)

To find the truck's original speed, we will use the following equation

$x=\frac{1}{2}(v_i+v_f)t$

Where x is the distance covered, vi is the initial speed, vf is the final speed, and t is the time. Solving for vi, we get:

$\begin{gathered} 2x=(v_i+v_f)t \\ \\ \frac{2x}{t}=v_i+v_f \\ \\ \frac{2x}{t}-v_f=v_i \end{gathered}$

Now, we can replace x = 40.0 m, t = 8.55s and vf = 2.45 m/s, so

$\begin{gathered} v_i=\frac{2(40.0\text{ m\rparen}}{8.55\text{ s}}-2.45\text{ m/s} \\ \\ v_i=6.91\text{ m/s} \end{gathered}$

Part (b)

Now, we can calculate the acceleration using the following equation:

$\begin{gathered} a=\frac{v_f-v_i}{t} \\ \\ a=\frac{2.45\text{ m/s-6.91 m/s}}{8.55\text{ s}} \\ \\ a=-0.52\text{ m/s}^2 \end{gathered}$

Therefore, the answers are

(a) 6.91 m/s

(b) -0.52 m/s²