Question

A point $\mathrm{P}$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $\mathrm{P}$ is such that it sweeps out a length $\mathrm{s}={\mathrm{t}}^3+5$ , where $\mathrm{s}$ is in meters and t is in seconds. The radius of the path is $20\mathrm{m}$ . The acceleration of $\mathrm{P}$ when $\mathrm{t}=2\mathrm{s}$ is nearly.

Answer 1

Step 1. Given

Length, $\mathrm{s}={\mathrm{t}}^3+5$

Radius, $\mathrm{r}=20\mathrm{m}$

Time, $\mathrm{t}=2\mathrm{s}$

We have to find net acceleration.

Step 2. Formula to be used

We get velocity from the first derivative of length with respect to time, and acceleration from the second derivative of length with respect to time.

Therefore, speed is,

$\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}$

$\mathrm{v}=3{\mathrm{t}}^2$

The rate of change of speed is,

${\mathrm{a}}_{\mathrm{t}}=\frac{\mathrm{dv}}{\mathrm{dt}}$

${\mathrm{a}}_{\mathrm{t}}=6\mathrm{t}$

We have to find net acceleration.

Step 3. Find the acceleration

There is a distinction between translational and centripetal acceleration.

So,

Tangent acceleration at $\mathrm{t}=2\mathrm{s}$ is,

${\mathrm{a}}_{\mathrm{t}}=6\times 2$

${\mathrm{a}}_{\mathrm{t}}=12\mathrm{m}/{\mathrm{s}}^2$

Now, at $\mathrm{t}=2\mathrm{s}$,

$\mathrm{v}=3{\mathrm{t}}^2$

$=3{\left(2\right)}^2$

$=12\mathrm{m}/\mathrm{s}$

Therefore, the centripetal acceleration is,

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{r}$

$=\frac{144}{20}\mathrm{m}/{\mathrm{s}}^2$

Step 4. Find the net acceleration

Now we will find the net acceleration.

So,

$\mathrm{a}=\sqrt{{\mathrm{a}}_{\mathrm{t}}^2+{\mathrm{a}}_{\mathrm{c}}^2}$

$=\sqrt{{\left(12\right)}^2+{\left(\frac{144}{20}\right)}^2}$

$=\sqrt{144+{\left(7.2\right)}^2}$

$=\sqrt{195.,84}$

$=\frac{12\sqrt{34}}{5}$

$=13.9942$

Hence, the net acceleration is equivalent to $14\mathrm{m}/{\mathrm{s}}^2$.