Question

159. A particle is moving in a straight line and passes
through a point O with a velocity of 6 m/s. The
particle moves with a constant retardation of 2
m/s2 for 4 s and there after moves with constant
velocity. How long after leaving o does the
particle return to O is:
(a) 3s
(b) 85
(c) 5s
(d) 10s​

Answer 1

Answer:

Option (b) - 8 s

Explanation:

u= 6 m/s

t= 4 s

a= -2 m/s²

s=ut+1/2 at²

s= 6*4 + 1/2*-2*4²

s= 24+1/2*-2*16

s= 24-16

s= 8 m

v= u+at

v= 6+ (-2*4)

v= 6-8

v= -2 m/s  ----------(i)

Now, since the particle returns to O on a straight line, displacement gets negative and v (final velocity) becomes u (initial velocity).

Since the particle moves with constant velocity while returning to O,

Time taken = Displacement / Velocity

Displacement = -8 m   [because particle goes in opposite direction]

Velocity= -2 m/s    [from (i) ]

Therefore, time taken to return to O= -8/-2 = 4 s

Total time taken (after leaving point O) = 4 s + 4 s

= 8 s.

Hope it helps!!!

Answer 2

Explanation:

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