1. Given data

• Radius of circular path: $r = 600\,\text{m}$
• Initial speed: $u = 54\,\text{km/hr} = 15\,\text{m/s}$
• Tangential acceleration magnitude equals centripetal acceleration magnitude.

So at every instant, $a_t = a_c$

2. Write expressions for the accelerations

• Tangential acceleration: $a_t = \frac{dv}{dt}$
• Centripetal acceleration: $a_c = \frac{v^2}{r}$

Given they are equal in magnitude, $\frac{dv}{dt} = \frac{v^2}{r}$

Substitute $r=600$: $\frac{dv}{dt} = \frac{v^2}{600}$

3. Relate speed and angular displacement

Since motion is along a circle, $v = r\frac{d\theta}{dt}$ So, $\frac{d\theta}{dt} = \frac{v}{r}$

Now, $\frac{dv}{d\theta} = \frac{dv/dt}{d\theta/dt} = \frac{v^2/r}{v/r} = v$

Hence, $\frac{dv}{d\theta} = v$

4. Solve for $v$ as a function of $\theta$

$\frac{dv}{v} = d\theta$ Integrating, $\ln v = \theta + C$

At $\theta=0$, $v=u=15$ m/s, so $C = \ln 15$ Thus, $v = 15e^{\theta}$

5. Find time for first quarter revolution

For first quarter revolution, $\theta = \frac{\pi}{2}$

Using $\frac{d\theta}{dt} = \frac{v}{r} = \frac{15e^{\theta}}{600} = \frac{e^{\theta}}{40}$ So, $dt = 40e^{-\theta} d\theta$

Integrate from $\theta=0$ to $\theta=\pi/2$: $T = \int_0^{\pi/2} 40e^{-\theta} d\theta$ $T = 40\left[-e^{-\theta}\right]_0^{\pi/2}$ $T = 40\left(1 - e^{-\pi/2}\right)$

This matches the given form: $T = t\left(1-e^{-\pi/2}\right)\,\text{s}$ Therefore, $t=40$

6. Comparison with stored answer

Derived answer: $40$

Stored correct answer: $40$

They match.