A body starts with velocity u and moves in a straight line with constant acceleration a. When it's velocity has increased to 5u, the acceleration is reversed in direction, the magnitude remains the same. Find the velocity with which the body passes the starting point again.
A body starts with velocity u and moves in a straight line with constant acceleration a. When it's velocity has increased to 5u, the acceleration is reversed in direction, the magnitude remains the same. Find the velocity with which the body passes the starting point again.
In the first case,
initial velocity = u,
Final velocity = v = 5u,
Acceleration = a
Using the third equaton of motion we have,
v²=u²+2as
2as = v²-u² = 25u²-u² = 24u²
s = (12u²)/a
Now the acceleration is changed to -a. Then the particle continues for time t in the same direction until the velocity becomes zero. Then the particle changes the direction and starts accelerating and passes over the point of start.
Initial velocity = 5u
Final velocity = v = 0
Acceleration = -a
Using the first equaton of motion we have,
v = 5u - at
=> 0 = 5u - at1
=> t = 5u/a.
Also, it takes t more time to stop and reverse direction.
The distance traveled/displacement in this time by using
s = u t + 1/2 a t²
=> s = 5u x (5u/a) - 1/2 x a(5u/a)² = 25u²/a - 25u²/2a = 25u²/2a
The total displacement from the initial point : 12u²/a + 25u²/2a = 49u²/2a
Now, acceleration = -a
u = 0
s = - 49u²/2a in the negative direction
Now applying, v²=u²+2as
v² = 0 - 2a(-49u²/2a ) = 49u²
v = 7u.
In the first case,
initial velocity = u,
Final velocity = v = 5u,
Acceleration = a
Using the third equaton of motion we have,
v²=u²+2as
2as = v²-u² = 25u²-u² = 24u²
s = (12u²)/a
Now the acceleration is changed to -a. Then the particle continues for time t in the same direction until the velocity becomes zero. Then the particle changes the direction and starts accelerating and passes over the point of start.
Initial velocity = 5u
Final velocity = v = 0
Acceleration = -a
Using the first equaton of motion we have,
v = 5u - at
=> 0 = 5u - at1
=> t = 5u/a.
Also, it takes t more time to stop and reverse direction.
The distance traveled/displacement in this time by using
s = u t + 1/2 a t²
=> s = 5u x (5u/a) - 1/2 x a(5u/a)² = 25u²/a - 25u²/2a = 25u²/2a
The total displacement from the initial point : 12u²/a + 25u²/2a = 49u²/2a
Now, acceleration = -a
u = 0
s = - 49u²/2a in the negative direction
Now applying, v²=u²+2as
v² = 0 - 2a(-49u²/2a ) = 49u²
v = 7u.
