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?
Answers
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
1. Reiterate the question in first sentence