Question

A particle starting with certain initial velocity and uniform acceleration covers a distance of 12 m in first 3 seconds and a distance of 30 m in next 3 seconds. The initial velocity of the particle is

• A. 3 ms\(^{-1}\)
• B. 2.5 ms\(^{-1}\)
• C. 2 ms\(^{-1}\)
• D. 1.5 ms\(^{-1}\)
• E. 1 ms\(^{-1}\)

Hint:

Always use the total distance for the second equation (12 + 30 = 42 m) at time \( t=6 \) s. It is much easier than calculating the velocity at \( t=3 \) s to use as the starting point for the second interval.

Answer 1

Answer: (E)

Concept: This problem utilizes the equations of motion for constant acceleration to solve for unknown initial conditions.
• Equation: \( s = ut + \frac{1}{2} at^2 \).

Step 1: Formulate equations for the two time periods.
1. For the first 3 seconds (\( s = 12 \)): \[ 12 = 3u + \frac{1}{2} a(3)^2 \implies 12 = 3u + 4.5a \implies 4 = u + 1.5a \quad \dots(1) \] 2. For the total 6 seconds (\( s = 12 + 30 = 42 \)): \[ 42 = 6u + \frac{1}{2} a(6)^2 \implies 42 = 6u + 18a \implies 7 = u + 3a \quad \dots(2) \]

Step 2: Solve the simultaneous equations.
Subtract equation (1) from equation (2): \[ (u + 3a) - (u + 1.5a) = 7 - 4 \] \[ 1.5a = 3 \implies a = 2 \text{ ms}^{-2} \] Substitute \( a = 2 \) back into equation (1): \[ 4 = u + 1.5(2) \] \[ 4 = u + 3 \implies u = 1 \text{ ms}^{-1} \]