Question

(c) 18m -
(d) 24m
142. A student is standing at a distance of 50 metres from the bus.
As soon as the bus begins its motion with an acceleration of
lms, 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
(6) 8 ms-1
te) 10 ms-1
(d) 12 ms -1
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Answer 1

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Answer:

u min = 10 m/s

EXPLANATION-

Acceleration of bus a = 1 m/s

In 10 seconds, the distance travelled by te bus S can be given as

S = ut + 1/2 at²

as initial velocity of bus is zero,

S = 0 + 0.5×1×10²

∴S = 50 m ....(1)

Student in behind the bus by 50 m

As calculated above, bus travels distance = 50 m.

Thus, the students must travel min distance = 50 + 50 = 100 m

Uniform velocity of student can be calculated as

U = Distance traveleed ÷ time taken

∴ U = 100 ÷ 10

∴ U = 10 m/s

Minimum value of u, (uniform velocity by the student) so that the student is able to catch the bus = 10 m/s