Acceleration a of a particle moving in a straight line varies with its velocity v (in m/s) as a = 2(2-v) m/s². Find the maximum possible speed of the particle, assuming at t = 0 particle was at rest.
Answers
Answer:
Let the distance between A and B=s and as mentioned C is the midpoint of A and B thus distance between A C and
C B=
2
s
From equation of motion we get distance between AB
v
2
=u
2
+2as
s=
2a
v
2
−u
2
----------(1)
Let speed of the vehicle be pm/s at point C
Again from equation of motion we get,
p
2
=u
2
+2a
2
s
as distance between A and C is
2
s
s=
a
p
2
−u
2
-------(2)
Computing 1 and 2 we get
a
p
2
−u
2
=
2a
v
2
−u
2
p
2
=
2
v
2
+u
2
p=
2
v
2
+u
2
Mark it as brainliest answer
Given : Acceleration a of a particle moving in a straight line varies with its velocity v (in m/s) as a = 2(2-v) m/s².
To Find : the maximum possible speed of the particle, assuming at t = 0 particle was at rest.
Solution:
a = 2(2-v)
Acceleration a of a particle moving in a straight line varies with its velocity v
Velocity will be max when a = 0
0 = 2(2 - v)
=> v = 2
Hence maximum possible speed of the particle is 2 m/s
Another method :
v = u + at
u = 0
a = 2(2-v)
v = 2(2-v)t
=> v = = 4t - 2vt
=> v(1 + 2t) = 4t
=> v = 4t/(1 + 2t)
dv/dt = -4t (2)/(1 + 2t)² + 4/(1 + 2t)
= (-8t + 4 + 8t)/( 1 + 2t)²
= 4/(1 + 2t)²
so dv/dt = 0 when t tends to infinity
d²v/dt² is - ve for t tends to infinity
Hence velocity
v = 4t/(1 + 2t)
= 4/( 1/t + 2)
as tends to infinity 1/t = 0
= 4 /(0 + 2)
= 2
Hence maximum possible speed of the particle is 2 m/s
Learn More:
A particle is dropped from a height of 80m . Find the distance ...
brainly.in/question/11370407
A ball is dropped from a height of 80ft the ball is such that it ...
brainly.in/question/11471174