Question

A truck covers 40.0 m in 8.20 s while uniformly slowing down to a final velocity of 3.50 m/s.
(a) Find the truck's original speed.
(b) Find its acceleration.

Answer 1

Answer on Question#34597 – Physics – Mechanics | Relativity

A truck covers $40.0\,\mathrm{m}$ in 8.20 s while uniformly slowing down to a final velocity of $3.50\,\mathrm{m/s}$.

(a) Find the truck's original speed.

(b) Find its acceleration.

Solution

The law of motion is

$$\begin{aligned} s(t) &= v_0 t + \frac{a t^2}{2} \\ v(t) &= v_0 + a t \end{aligned}$$

If $t = 8.2\,\mathrm{s}$

$$\begin{aligned} s(t) &= 40.0\,\mathrm{m} \\ v(t) &= 3.50\,\mathrm{m/s} \end{aligned}$$ $$\begin{cases} 40.0\,\mathrm{m} = v_0 \cdot 8.2 + \dfrac{a \cdot (8.2\,\mathrm{s})^2}{2} \\ 3.5\,\mathrm{m/s} = v_0 + a \cdot 8.2\,\mathrm{s} \end{cases} \Rightarrow$$ $$(40.0 - 28.7) = -a \cdot 33.62\,\mathrm{s^2}$$ $$a \approx -0.336\,\mathrm{m/s^2}$$ $$v_0 \approx 6.256\,\mathrm{m/s}$$

Answer:

Acceleration and initial speed are

$$\begin{aligned} a \approx -0.336\,\mathrm{m/s^2} \\ v_0 \approx 6.256\,\mathrm{m/s} \end{aligned}$$

Answer provided by www.AssignmentExpert.com