Question

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of $400\mathrm{m}$ in $20\mathrm{s}$. Find its acceleration. Find the force acting on it if its mass is $7\mathrm{tonnes}$.

Answer 1

Step 1: Given that:

Initial velocity($\mathrm{u}$) of the truck= $0$

Distance($\mathrm{s}$) travelled by the truck= $400\mathrm{m}$

Time($\mathrm{t}$) taken = $20\mathrm{s}$

Mass($\mathrm{m}$) of the truck = $7\mathrm{tonnes}$= $7\times 1000\mathrm{kg}=7000\mathrm{kg}$

Step 2: Formula used:

The second equation of motion is given as,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Where, $\mathrm{s}$ = distance covered by a body,$\mathrm{u}$= initial velocity of the body, $\mathrm{a}$= acceleration and $\mathrm{t}$= time.

And according to the second law of motion;

$\mathrm{Force}\left(\mathrm{F}\right)=\mathrm{Mass}\left(\mathrm{m}\right)\times \mathrm{Acceleration}\left(\mathrm{a}\right)$

Step 3: Calculation of acceleration($\mathrm{a}$) in the truck:

$$\begin{gathered} 400\mathrm{m}=0\times 20\mathrm{s}+\frac{1}{2}\times \mathrm{a}\times {\left(20\mathrm{s}\right)}^2 \\ \frac{1}{2}\times 400\mathrm{a}=400 \\ 400\mathrm{a}=400\times 2 \\ \mathrm{a}=\frac{800}{400} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$

Step 4: Calculation of force($\mathrm{F}$) on the truck:

$$\begin{gathered} \mathrm{F}=7000\mathrm{kg}\times 2\mathrm{m}/{\mathrm{s}}^2 \\ \mathbf{F}\mathbf{=}\mathbf{14000}\mathbf{N} \end{gathered}$$

Thus, the acceleration of the truck= $\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ and the force on the truck= $\mathbf{14000}\mathbf{N}$.