A particle start from rest with a velocity of 10 m / s and moves with a constant acceleration till the velocity increases to 100 m / s. At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

Question

A particle start from rest with a velocity of 10 m / s and moves with a constant acceleration till the velocity increases to 100 m / s. At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point? (A) 10 m / s (B) 20 m / s (C) 30 m / s (D) 40 m / s

Tardigrade Admin · Accepted Answer

According to relation = v 2- u 2=2 as (100)2-(10)2 =2 as 10000-100 =2 as 9900 =2 as Now acceleration just reversed = a =- a Particle comes back to original position v=? v 2- u 2=2 as (v)2-(100)2=2(-a) s v2=10000-2 as v2=10000-2 as Put 2as =9900 v 2=100 v =√100=10 m / s

Q. A particle start from rest with a velocity of 10m/s and moves with a constant acceleration till the velocity increases to 100m/s. At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

10 m / s

20 m / s

30 m / s

40 m / s

According to relation =v2−u2=2as (100)2−(10)2=2as 10000−100=2as 9900=2as Now acceleration just reversed =a=−a Particle comes back to original position v=? v2−u2=2 as (v)2−(100)2=2(−a)s v2=10000−2as v2=10000−2as Put 2as =9900 v2=100 v=100​=10m/s

Q. A particle start from rest with a velocity of $10 m / s$ and moves with a constant acceleration till the velocity increases to $100 m / s$ . At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?