1. Given data

• Arc length covered by the particle: $s=t^3+5$
• Radius of circular path: $r=20\text{ m}$
• Need acceleration at: $t=2\text{ s}$

For motion on a circle, acceleration has two perpendicular components:

• Tangential acceleration: $a_t=\frac{dv}{dt}=\frac{d^2 s}{dt^2}$
• Normal (centripetal) acceleration: $a_n=\frac{v^2}{r}$

The total acceleration is: $a=\sqrt{a_t^2+a_n^2}$

2. Find speed

Since speed is rate of change of arc length, $v=\frac{ds}{dt}=\frac{d}{dt}(t^3+5)=3t^2$

At $t=2$ s, $v=3(2)^2=12\text{ m/s}$

3. Find tangential acceleration

$a_t=\frac{dv}{dt}=\frac{d}{dt}(3t^2)=6t$

At $t=2$ s, $a_t=6(2)=12\text{ m/s}^2$

4. Find normal acceleration

$a_n=\frac{v^2}{r}=\frac{12^2}{20}=\frac{144}{20}=7.2\text{ m/s}^2$

5. Find resultant acceleration

$a=\sqrt{a_t^2+a_n^2}$

$a=\sqrt{12^2+7.2^2}$

$a=\sqrt{144+51.84}$

$a=\sqrt{195.84}\approx 13.99\text{ m/s}^2$

So the acceleration is nearly $14\text{ m/s}^2$

6. Option check

• A: $13\text{ m/s}^2$ ❌
• B: $12\text{ m/s}^2$ ❌
• C: $7.2\text{ m/s}^2$ ❌
• D: $14\text{ m/s}^2$ ✅

Therefore, the correct answer is D.