The problem given asks us to find the exact value of $T$ when a particle $P$ comes to instantaneous rest.

Given:

• Displacement of the particle: $s = 5t^2 - t^3$
• Time: $0 \leq t \leq 5$
• We need to find $T$ such that $P$ comes to instantaneous rest at $t = T$ (i.e., when velocity $v = 0$).

Solution:

1. Find velocity: Velocity is the derivative of displacement with respect to time. $v(t) = \frac{ds}{dt} = \frac{d}{dt}(5t^2 - t^3)$ Differentiate: $v(t) = 10t - 3t^2$

2. Find when velocity is zero (instantaneous rest): $v(T) = 0 \quad \Rightarrow \quad 10T - 3T^2 = 0$ Factor the equation: $T(10 - 3T) = 0$ So, the solutions are: $T = 0 \quad \text{or} \quad T = \frac{10}{3}$

Since $T > 0$, we take $T = \frac{10}{3}$ seconds.

Conclusion:

The exact value of $T$ is $\frac{10}{3}$ seconds.

Would you like further details on any step, or do you have more questions?

Here are 5 related questions to deepen your understanding:

1. How do you find the displacement at $T = \frac{10}{3}$ seconds?
2. What is the acceleration of the particle at $t = T$?
3. Can the particle's velocity be negative at any point during $0 \leq t \leq 5$?
4. How do you determine the total distance traveled by the particle over $t = 0$ to $t = 5$?
5. What does the shape of the displacement-time graph look like for this function?

Tip:

Remember, the point of instantaneous rest corresponds to where velocity is zero, but acceleration may still act on the particle, influencing future motion!