Question

A particle starts from rest and moves at an acceleration of 4m/s² for 10 seconds and then decelerates uniformly for another 5 seconds to come to rest. The distance covered by the particle in this journey is:

A) 100 meters

B) 150 meters

C) 200 meters

D) 250 meters

Answer 1

Using kinematic equations, the particle covers 200 meters during the acceleration phase and 100 meters during the deceleration phase, totaling 300 meters for the entire journey, which is not listed in the provided options.

Step-by-step explanation:

The distance covered by a particle that starts from rest, accelerates at 4 m/s² for 10 seconds, and then decelerates uniformly to come to a rest in another 5 seconds can be calculated using the kinematic equations of motion. During the acceleration phase, the final velocity (v) after 10 seconds can be found using the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Since the particle starts from rest, u = 0, so v = 0 + (4 m/s²)(10 s) = 40 m/s. The distance (s) covered in the first 10 seconds can be found using s = ut + ½at², resulting in s = 0 + ½(4 m/s²)(10 s)² = 200 m.

When decelerating, the particle goes from 40 m/s to 0 m/s in 5 seconds.

This requires a deceleration of a = (0 - 40 m/s) / 5 s = -8 m/s².

Using the equation s = ut + ½at² again,

the distance covered during deceleration is s = (40 m/s)(5 s) + ½(-8 m/s²)(5 s)^2 = 100 m.

Summing the distances from both phases, the total distance covered by the particle is 300 meters, which is not one of the options provided.