Question

A student is at a distance of 16 m from a bus. The bus starts accelerating from rest with $ a = 9 \, \text{m/s}^2 $. The minimum velocity with which the student should run towards the bus to catch it is $ \alpha \sqrt{2} \, \text{m/s} $. Find $ \alpha $.

• A. 10
• B. 12
• C. 15
• D. 20

Hint:

Use the concept of relative motion and ensure time taken is equal. Apply discriminant zero condition for minimum velocity problems.

Answer 1

The Correct Option is B

Let the student run with constant speed \( v = \alpha \sqrt{2} \), and the bus starts from rest with acceleration \( a = 9 \, \text{m/s}^2 \). Let the time taken to meet be \( t \). - Distance travelled by student: \[ x_s = v t = \alpha \sqrt{2} t \] - Distance travelled by bus (starting from rest): \[ x_b = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 9 t^2 = \frac{9}{2} t^2 \] Since initial separation is 16 m: \[ x_s = x_b + 16 \Rightarrow \alpha \sqrt{2} t = \frac{9}{2} t^2 + 16 \] Multiply both sides by 2: \[ 2\alpha \sqrt{2} t = 9t^2 + 32 \Rightarrow 9t^2 - 2\alpha \sqrt{2} t + 32 = 0 \] For student to just catch the bus, the equation should have only one solution ⇒ Discriminant \( D = 0 \) \[ D = (2\alpha \sqrt{2})^2 - 4 \cdot 9 \cdot 32 = 0 \Rightarrow 4\alpha^2 \cdot 2 = 1152 \Rightarrow 8\alpha^2 = 1152 \Rightarrow \alpha^2 = 144 \Rightarrow \alpha = 12 \]