Question:
A passenger sees a bus standing 10 m ahead of him. The bus starts moving with a constant acceleration of \(5 \, m\,s^{-2}\) away from him. The passenger runs with constant speed to catch the bus. The minimum speed required to catch the bus is:
A passenger sees a bus standing 10 m ahead of him. The bus starts moving with a constant acceleration of \(5 \, m\,s^{-2}\) away from him. The passenger runs with constant speed to catch the bus. The minimum speed required to catch the bus is:
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For catch-up problems, always equate positions and apply the discriminant condition for minimum required speed.
Updated On: Jun 19, 2026
- \(10\sqrt{2}\, m\,s^{-1}\)
- \(5\sqrt{2}\, m\,s^{-1}\)
- \(10\, m\,s^{-1}\)
- \(50\, m\,s^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Define motion of bus.
Let the bus start from rest at \(t=0\), 10 m ahead of the passenger. It moves with constant acceleration \(a = 5 \, m\,s^{-2}\).
So displacement of bus is: \[ x_b = 10 + \frac{1}{2}at^2 = 10 + 2.5t^2 \]
Step 2: Define motion of passenger.
The passenger runs with constant speed \(u\). His displacement is: \[ x_p = ut \]
Step 3: Condition for catching the bus.
For catching, positions must be equal: \[ ut = 10 + 2.5t^2 \]
Rearranging: \[ 2.5t^2 - ut + 10 = 0 \]
Step 4: Condition for real solution.
For the passenger to catch the bus, this quadratic equation must have real roots. So discriminant must satisfy: \[ u^2 - 4(2.5)(10) \geq 0 \]
Step 5: Solve inequality.
\[ u^2 - 100 \geq 0 \Rightarrow u^2 \geq 100 \Rightarrow u \geq 10 \, m\,s^{-1} \]
Step 6: Final conclusion.
Thus, the minimum speed required is \(10 \, m\,s^{-1}\).
Final Answer: \[ \boxed{10 \, m\,s^{-1}} \]
Let the bus start from rest at \(t=0\), 10 m ahead of the passenger. It moves with constant acceleration \(a = 5 \, m\,s^{-2}\).
So displacement of bus is: \[ x_b = 10 + \frac{1}{2}at^2 = 10 + 2.5t^2 \]
Step 2: Define motion of passenger.
The passenger runs with constant speed \(u\). His displacement is: \[ x_p = ut \]
Step 3: Condition for catching the bus.
For catching, positions must be equal: \[ ut = 10 + 2.5t^2 \]
Rearranging: \[ 2.5t^2 - ut + 10 = 0 \]
Step 4: Condition for real solution.
For the passenger to catch the bus, this quadratic equation must have real roots. So discriminant must satisfy: \[ u^2 - 4(2.5)(10) \geq 0 \]
Step 5: Solve inequality.
\[ u^2 - 100 \geq 0 \Rightarrow u^2 \geq 100 \Rightarrow u \geq 10 \, m\,s^{-1} \]
Step 6: Final conclusion.
Thus, the minimum speed required is \(10 \, m\,s^{-1}\).
Final Answer: \[ \boxed{10 \, m\,s^{-1}} \]
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