## Physics Kinematics- Acceleration and velocity

A student is standing 50 m behind a bus. As soon as the bus begins its motion with acceleration 1 ms-2, the student starts running towards the bus wit a constant velocity u. Assuming the motion to be along a staright road, what is the minimum value of u, so that the student is able to catch the bus?
1. 5 ms-1
2. 8 ms-1
3. 10 ms-1
4. 12 ms-1

18-Jun-2016 5:41 PM
~6

0

The bus and the student both start moving.

Suppose at time t, the student catches the bus. Thus, the bus would have covered x distance in that time, and the student is supposed to cover (x+50) distance in the same time t.

The bus covered x distance from initial position,

$x = 0.5 \times 1 \times t^2$

$x = \frac{t^2}{2}$  ----(1)

For student,

$(x + 50) = u \times t$........................................(2)

substituute eq. (1) from eq. (2)

$t^2/2 + 50 = ut$

$u = t/2 + 50/t$

$du/dt = 1/2 - 50/t^2 = 0$ (since student is moving with const velocity)

$50/t^2 = 0.5$

$t^2 = 50/0.5 = 100$

t = 10 seconds

Therefore,

u = 10/2 + 50/10

= 5 + 5

= 10 m/s

20-06-2016 19:23
~4