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
Given: The initial velocity of a particle moving along a straight line is 5 m/s and its retardation is 1 ms^-2.
To find: The distance moved by the particle in fourth second of its motion?
Solution:
- Now we have given :
u = 5 m/s
a = -1 ms^-2
n = 4
- We know the formula of distance travelled in nth second, which is:
Sn = u + a(2n-1)/2
S = 5 + (-1)(2(4) - 1)/2
S = 5 + (-1)(8 - 1)/2
S = 5 + (-1)(7)/2
S = 5 - 7/2
S = 10 - 7 / 2
S = 3/2
Answer:
So the distance travelled in 4th second is 3/2 m.
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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