Question

A particle moves for 50 seconds if first accelerates from rest and then retard or deaccelerates to rest. If the retardation be 5 times the acceleration, then the time for retardation is:

• A. 25/3 second
• B. 50/3 second
• C. 25 second
• D. 100/3 second

Answer 1

Answer: (A)

25/3 second

Answer 2

The problem involves two phases of motion: acceleration from rest and deceleration to rest. We can use the equations of motion for constant acceleration.

Let:

• a be the acceleration during the first phase.
• t_a be the time for acceleration.
• a_r be the retardation (deceleration) during the second phase.
• t_r be the time for retardation.
• v_max be the maximum velocity attained by the particle (at the end of acceleration phase and beginning of retardation phase).

Phase 1: Acceleration The particle starts from rest (initial velocity u = 0) and accelerates to v_max. Using the first equation of motion, v = u + at: v_max = 0 + a * t_a v_max = a * t_a (Equation 1)

Phase 2: Retardation The particle starts with velocity v_max and decelerates to rest (final velocity v = 0). The retardation a_r is given as 5 times the acceleration a, so a_r = 5a. Using the first equation of motion, v = u + at (here a is -a_r for deceleration): 0 = v_max + (-a_r) * t_r 0 = v_max - (5a) * t_r v_max = 5a * t_r (Equation 2)

Equating v_max from both phases: From Equation 1 and Equation 2: a * t_a = 5a * t_r Since a is non-zero (the particle accelerates), we can divide both sides by a: t_a = 5 * t_r (Equation 3)

Total Time: The total time for which the particle moves is given as 50 seconds: t_a + t_r = 50 (Equation 4)

Solving the system of equations: Substitute t_a from Equation 3 into Equation 4: (5 * t_r) + t_r = 50 6 * t_r = 50 t_r = 50 / 6 t_r = 25 / 3 seconds

Thus, the time for retardation is 25/3 seconds.

The final answer is $\boxed{\text{25/3 second}}$.