Question:
A student is at a distance \(16\,\text{m}\) from a bus when the bus begins to move with a constant acceleration of \(9\,\text{m s}^{-2}\). The minimum velocity with which the student should run towards the bus so as to catch it is \(\alpha\sqrt{2}\,\text{m s}^{-1}\). The value of \(\alpha\) is
A student is at a distance \(16\,\text{m}\) from a bus when the bus begins to move with a constant acceleration of \(9\,\text{m s}^{-2}\). The minimum velocity with which the student should run towards the bus so as to catch it is \(\alpha\sqrt{2}\,\text{m s}^{-1}\). The value of \(\alpha\) is
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For minimum speed problems involving catching a moving object, form a quadratic in time and use the condition \(D=0\).
Updated On: Jun 22, 2026
- \(10\)
- \(12\)
- \(15\)
- \(20\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the positions of bus and student.
Let the initial position of the student be \[ x=0 \] The bus is initially \(16\,\text{m}\) ahead of the student.
Since the bus starts from rest with acceleration \[ a=9\,\text{m s}^{-2}, \] its displacement after time \(t\) is \[ s_b=16+\frac{1}{2}at^2 \] \[ s_b=16+\frac{9}{2}t^2 \] If the student runs with constant velocity \(v\), then his displacement is \[ s_s=vt \]
Step 2: Apply catching condition.
For the student to catch the bus, \[ vt=16+\frac{9}{2}t^2 \] Rearranging, \[ \frac{9}{2}t^2-vt+16=0 \] For the minimum velocity, this quadratic in \(t\) must have equal roots.
So, \[ D=0 \]
Step 3: Use discriminant condition.
\[ (-v)^2-4\left(\frac{9}{2}\right)(16)=0 \] \[ v^2-288=0 \] \[ v^2=288 \] \[ v=\sqrt{288} \] \[ v=12\sqrt{2} \] Given, \[ v=\alpha\sqrt{2} \] Hence, \[ \alpha=12 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{12} \]
Let the initial position of the student be \[ x=0 \] The bus is initially \(16\,\text{m}\) ahead of the student.
Since the bus starts from rest with acceleration \[ a=9\,\text{m s}^{-2}, \] its displacement after time \(t\) is \[ s_b=16+\frac{1}{2}at^2 \] \[ s_b=16+\frac{9}{2}t^2 \] If the student runs with constant velocity \(v\), then his displacement is \[ s_s=vt \]
Step 2: Apply catching condition.
For the student to catch the bus, \[ vt=16+\frac{9}{2}t^2 \] Rearranging, \[ \frac{9}{2}t^2-vt+16=0 \] For the minimum velocity, this quadratic in \(t\) must have equal roots.
So, \[ D=0 \]
Step 3: Use discriminant condition.
\[ (-v)^2-4\left(\frac{9}{2}\right)(16)=0 \] \[ v^2-288=0 \] \[ v^2=288 \] \[ v=\sqrt{288} \] \[ v=12\sqrt{2} \] Given, \[ v=\alpha\sqrt{2} \] Hence, \[ \alpha=12 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{12} \]
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