Question

A ball starts moving eastward with speed 35 m/s. An acceleration of 10m/s2 is acting on it in westward direction. What is distance travelled by particle in 4th second of its motion?

Answer 1

A ball starts moving eastward with speed 35 m/s. An acceleration of 10m/s2 is acting on it in westward direction. What is distance travelled by particle in 4th second of its motion?

Initial Velocity u = 35 m/s  Eastward

acceleration a = - 10 m/s² eastward ( as acceleration was given westward so it will be actually retardation so - sign added)

S = ut + (1/2)at²

Distance travelled in 3 sec

S₃ = 35*3 +(1/2)(-10)3²

S₃ = 105 - 45

S₃ = 60 m

Distance traveled in 4 secs =

S₄ = 35*4 + (1/2)(-10)4²

=>S₄ = 140 - 80

=> S₄ = 60 m

Distance traveled in 4th Sec =  S₄  - S₃ = 60 - 60 = 0 m

it is actually displacement . it means direction of ball has changed in between

Lets find out at what moment speed become zero

v = u + at

0 = 35 - 10t

t = 3.5 sec

at 3.5 sec ball comes to rest and then change direction

Distance traveled in 3.5 sec

S = 35*3.5 + (1/2)(-10)3.5²

S = 122.5 - 61.25

S = 61.25 m

Distance traveled from 3 to 3.5 sec = 61.25 - 60 = 1.25 m

Distance traveled in next 0.5 sec , Considering from 3.5 sec

speed at t= 3.5 = 0   (4 sec from initial = 4-3.5 = 0.5 sec from 3.5 sec)

acceleration is taken positive as we are calculation distance and not displacement

S = 0 + (1/2)(10)(0.5)²

S = 1.25 m

total Distance covered in 4th Sec = 1.25 + 1.25 = 2.5 m