Question

A particle starts moving from rest under a constant acceleration, and covers a distance ‘X’ in ‘t’ seconds. The distance covered in next ‘t’ seconds is

• A. X
• B. 2X
• C. 3X
• D. 4X

Hint:

When motion starts from rest under constant acceleration: Distance in first \(t\) seconds = \(X\), Distance in next \(t\) seconds = \(3X\)

Answer 1

The Correct Option is C

The displacement of a particle moving with constant acceleration from rest is given by: \[ S = ut + \frac{1}{2}at^2 \] Since initial velocity \(u = 0\), the distance covered in time \(t\) is: \[ X = \frac{1}{2}at^2 \quad \text{(Equation 1)} \] Now, total distance covered in time \(2t\) is: \[ S_{2t} = \frac{1}{2}a(2t)^2 = \frac{1}{2}a \cdot 4t^2 = 2at^2 \] Distance covered in the next \(t\) seconds = \(S_{2t} - S_t\) \[ = 2at^2 - \frac{1}{2}at^2 = \frac{3}{2}at^2 = 3X \quad \text{(from Equation 1)} \] Hence, the particle travels 3X distance in the next \(t\) seconds.

Answer 2

Step 1: Understand the given information
- Initial velocity, \(u = 0\) (starts from rest)
- Constant acceleration, \(a\)
- Distance covered in first \(t\) seconds is \(X\)

Step 2: Use the equation of motion to find \(X\)
\[ X = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a t^2 = \frac{1}{2} a t^2 \]

Step 3: Find distance covered in next \(t\) seconds (from \(t\) to \(2t\))
Total distance covered in \(2t\) seconds:
\[ S_{2t} = ut' + \frac{1}{2} a (2t)^2 = 0 + \frac{1}{2} a (4 t^2) = 2 a t^2 \]

Distance covered in next \(t\) seconds = distance covered in \(2t\) seconds – distance covered in first \(t\) seconds:
\[ S = S_{2t} - X = 2 a t^2 - \frac{1}{2} a t^2 = \frac{3}{2} a t^2 \]

Step 4: Express the distance in terms of \(X\)
Recall \(X = \frac{1}{2} a t^2\), so:
\[ S = 3 \times \frac{1}{2} a t^2 = 3X \]

Step 5: Final Conclusion
The distance covered in the next \(t\) seconds is \(3X\).