A particle starts from the origin with a velocity of 10 ms and moves w...
Problem Statement
A particle starts from the origin with a velocity of 10 ms and moves with a constant acceleration till the velocity increase to 50 ms at that instant the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?
Solution
Step 1: Initial Velocity and Acceleration
Given, initial velocity (u) = 10 m/s
Let, acceleration (a) = a1 (constant)
Using the equation of motion, v = u + at, where v is the final velocity, and t is the time taken,
when the velocity increases to 50 m/s,
50 = 10 + a1t1
where t1 is the time taken for the velocity to increase to 50 m/s.
Step 2: Reversed Acceleration
After the velocity reaches 50 m/s, the acceleration is suddenly reversed. Therefore, the acceleration becomes -a1, and the velocity starts to decrease.
Using the equation of motion, v = u + at, where v is the final velocity, and t is the time taken,
0 = 50 - a1t2
where t2 is the time taken for the velocity to decrease to 0 m/s.
Step 3: Total Time of Travel
The total time of travel can be calculated as the sum of the time taken for the velocity to increase to 50 m/s and the time taken for the velocity to decrease to 0 m/s.
Therefore, the total time of travel, T = t1 + t2.
Step 4: Final Velocity
Using the equation of motion, v = u + at, where v is the final velocity, and t is the time taken,
when the velocity reaches the origin,
0 = 50 - a1t2 + (-a1)T
Substituting T = t1 + t2, we get
0 = 50 - a1t2 - a1(t1 + t2)
Simplifying, we get
0 = 50 - a1T
Therefore, the final velocity when the particle returns to the starting point is 10 m/s.
Conclusion
Therefore, the final velocity of the particle when it returns to the starting point is 10 m/s.