Question

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

Answer 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.

Answer 2

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

Learn More:

a particle is dropped from the top of a very high tower then the ...

https://brainly.in/question/10206693

A train starts from rest and accelerates uniformly at a rate of 2m/s^2 for

https://brainly.in/question/336565