A particle is travelling along X-axis and its x-coordinate is related to time as follows:
`x=5t^(2)-20`
Here x is measured in metres and time t in seconds.
When does the magnitude of velocity become equal to that of acceleration?

Options

A. 2s
B. 3s
C. 1s
D. never

Explanation

To solve the problem, we need to find the time when the magnitude of velocity becomes equal to that of acceleration for the given equation of motion \( x = 5t^2 - 20 \).

### Step-by-Step Solution:

1. **Identify the position function**: The position of the particle is given by the equation:
\[ x(t) = 5t^2 - 20 \]

2. **Find the velocity**: The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(5t^2 - 20) \]
Using the power rule of differentiation:
\[ v(t) = 10t \]

3. **Find the acceleration**: The acceleration \( a(t) \) is the derivative of the velocity function with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(10t) \]
Differentiating gives:
\[ a(t) = 10 \]

4. **Set the magnitudes equal**: We need to find when the magnitude of velocity equals the magnitude of acceleration:
\[ |v(t)| = |a(t)| \]
Since \( a(t) = 10 \) is a constant, we have:
\[ |10t| = 10 \]

5. **Solve for \( t \)**: This gives us two cases to consider:
- Case 1: \( 10t = 10 \)
- Case 2: \( 10t = -10 \)

Solving Case 1:
\[ t = 1 \text{ second} \]

Solving Case 2:
\[ t = -1 \text{ second (not physically meaningful in this context)} \]

6. **Conclusion**: The only valid solution is:
\[ t = 1 \text{ second} \]

### Final Answer:
The magnitude of velocity becomes equal to that of acceleration at \( t = 1 \) second.

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