The initial velocity of a particle moving along a straight line is 5 m/s and its retardation is 1 ms–2. The distance moved by the particle in fourth second of its motion is
Answers
Answered by
1
Answer:
Given,
u=10m/s
a=−2m/s
2
n=5
The distance covered by the particle in n
th
second is
S
n
=u+
2
a
(2n−1)
S
5
=10−
2
2
(2×5−1)
S
5
=10−9=1m
The correct option is A.
Answered by
5
3/2 m is the distance covered in 4th sec
Step-by-step explanation:
Initial velocity = 5 m/s
retardation is 1 ms-2.
=> acceleration = - 1 m/s²
V = U + at
Velocity after 3 secs = 5 + (-1)3 = 2 m/s
Velocity after 4 secs = 5 + (-1)4 = 1 m/s
using V² - U² = 2aS
=> 1² - 2² = 2(-1) S
=> 1 - 4 = - 2S
=> - 3 = -2S
=> S = 3/2
3/2 m is the distance covered in 4th sec
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