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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!

A particle P is moving along a straight line. At time t seconds, the displacement, s metres, of P from a fixed point O on the line is given by s = 5t^2 - t^3, 0 ≤ t ≤ 5. At time T seconds, where T > 0, P comes to instantaneous rest. Find the exact value of T.

Learn how to find the exact time when a particle comes to rest using calculus. The particle’s displacement is given as s = 5t^2 - t^3. This problem involves key concepts of differentiation and solving quadratic equations to find when velocity equals zero. Suitable for students in Grades 11-12.

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