Question

A particle is moving along a circular path of radius $0.7\mathrm{m}$ with tangential acceleration of $5{\mathrm{ms}}^{-2}$ The particle is initially at rest.
Based on above information, answer the following question:

3.The distance travelled by the particle in $7\text{s}$ is

• A. $123\mathrm{m}$
• B. $725\mathrm{m}$
• C. $728\mathrm{m}$
• D. $426\mathrm{m}$

Answer 1

The correct option is A

$123\mathrm{m}$

Step 1: Given data:

Tangential acceleration and radius : ${\mathrm{\alpha }}_{\mathrm{T}}=5{\mathrm{ms}}^{-2},\mathrm{R}=0.7\mathrm{m}$.

Formula :
Tangential acceleration, ${\mathrm{\alpha }}_{\mathrm{T}}=\mathrm{R\alpha }$ where $\mathrm{\alpha }=$angular acceleration.

Angular displacement is given as $\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2$ where ${\mathrm{\omega }}_0=$ initial angular velocity and here ${\mathrm{\omega }}_0=0$.

Step 2: To find angular acceleration:

Angular acceleration,
$$\begin{gathered} \mathrm{\alpha }=\frac{{\mathrm{\alpha }}_{\mathrm{T}}}{\mathrm{R}} \\ \Rightarrow \mathrm{\alpha }=\frac{5}{0.7} \\ \Rightarrow \mathrm{\alpha }=7.143{\mathrm{rads}}^{-2} \end{gathered}$$

Step 3: To find angular displacement:
We know that the angular displacement is given as
$\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2$
Where ${\mathrm{\omega }}_0=$ initial angular velocity and here ${\mathrm{\omega }}_0=0$
$$\begin{gathered} \therefore \mathrm{\theta }=\frac{1}{2}{\mathrm{\alpha t}}^2 \\ \Rightarrow \mathrm{\theta }=\frac{1}{2}*7.143*{\left(7\right)}^2\mathrm{rad} \\ \Rightarrow \mathrm{\theta }=175\mathrm{rad} \end{gathered}$$

Step 4: To find the linear displacement:

We know that the linear distance travelled is given as
$$\begin{gathered} \mathrm{S}=\mathrm{R\theta } \\ \Rightarrow \mathrm{S}=0.7*175 \\ \Rightarrow \mathrm{S}=122.5\mathrm{m} \\ \Rightarrow \mathrm{S}\approx 123\mathrm{m} \end{gathered}$$

The distance travelled by the particle in $7\text{s}$ is $123\mathrm{m}$.

Hence, option (a) is correct.