Question

the initial velocity of a particle is 10 M per second and its retardation is 20 m per second square the distance covered in the fifth second of the motion will be question

Answer 1

Initial velocity = 10 m/s
acceleration = -20 m/s²
time = 5 seconds
distance = ?

using second law of uniformly accelerated motion,
we have,
s = ut + 1/2at²
s = 50 + 1/2×-20×25
s = 50 - 250
s = -200 metres.----(1)

Distance covered till fifth second of motion will be -200 metres.
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taking time = 4 seconds.

using second equation of uniformly accelerated motion,
we have,
s = ut + 1/2at²
s = 40 + 1/2×-20×16
s = 40 - 160
s = -120 metres.
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Distance covered from 4th second to 5th second.
-200-(-120)
-200+120
-80

Answer 2

u = 10m/s²
at t = 4s
from eqn 1
v = u + at
v = 10 + ( -20 ) x 4 = 10 - 80
v1 = -70 m/s

now, by eqn 1

v = 10 + (-20) x 5
v2 = -90 m/s

now, normal speed in 5th second
v = ( v1-v2)/2
= ( -70 + 90 ) / 2
= 10 m/s

distance = speed x time
= 10 x 1 = 10 m

solved !