A student is standing at a distance of 50 metres from a bus. As soon as the bus begins its motion (starts moving away from student) with an acceleration of 1 ms-2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is:

Question

A student is standing at a distance of 50 metres from a bus. As soon as the bus begins its motion (starts moving away from student) with an acceleration of 1 ms-2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is: (A) 5 ms-1 (B) 8 ms-1 (C) 10 ms-1 (D) 12 ms-1

Tardigrade Admin · Accepted Answer

50 + x = ut ...(1) and x = (t2/2) ...(2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/431e77f44035e00f2b478af86bf343ef-.png" /> From (1) and (2) 50 + (t2/2) = ut ⇒ t2 - 2ut + 100 = 0 For real roots, discriminant = 0 For minimum velocity, discriminant = 0 ⇒ 4u2 -400 = 0 ⇒ u = 10 m/s

$5 m s^{- 1}$

$8 m s^{- 1}$

$10 m s^{- 1}$

$12 m s^{- 1}$

$50 + x = u t ... (1)$ and $x = \frac{t ^{2}}{2} ... (2)$

From $(1)$ and $(2)$

$50 + \frac{t ^{2}}{2} = u t$

$\Rightarrow t^{2} - 2 u t + 100 = 0$

For real roots, discriminant $= 0$

For minimum velocity, discriminant $= 0$

$\Rightarrow 4 u^{2} - 400 = 0$

$\Rightarrow u = 10 m / s$

5ms−1

8ms−1

10ms−1

12ms−1

50+x=ut...(1) and x=2t2​...(2) From (1) and (2) 50+2t2​=ut ⇒t2−2ut+100=0 For real roots, discriminant =0 For minimum velocity, discriminant =0 ⇒4u2−400=0 ⇒u=10m/s

$5 m s^{- 1}$

$8 m s^{- 1}$

$10 m s^{- 1}$

$12 m s^{- 1}$

$50 + x = u t ... (1)$ and $x = \frac{t ^{2}}{2} ... (2)$

From $(1)$ and $(2)$

$50 + \frac{t ^{2}}{2} = u t$

$\Rightarrow t^{2} - 2 u t + 100 = 0$

For real roots, discriminant $= 0$

For minimum velocity, discriminant $= 0$

$\Rightarrow 4 u^{2} - 400 = 0$

$\Rightarrow u = 10 m / s$