Question

A particle is moving with initial velocity 40m/s along positive X-axis. Acceleration is -10m/s2. It starts from x = 10 m. Find out maximum co-ordinate of particle in positive direction.
(1) 90 m
(2) O m
(3) 120 m
(4) 80 m


please answer!!!
ASAP!!​

Answer 1

Given :

Initial velocity , u = 40 m/s

Acceleration = - 10 m / s^2

Initial starting coordinate = 10 m

To find :

The maximum coordinate of the particle in positive direction.

Solution :

As the direction of the velocity and acceleration is opposite , particle will go to maximum distance till its velocity becomes zero .

Now ,

v^2 = u^2 + 2 * a * s

=> 0 = 40 * 40 - 2 * 10 * s

=> 20 * s = 1600

=> s = 80 m

The maximum coordinate of the particle in positive direction = 80 m + 10 m = 90 m

The maximum coordinate of the particle in positive direction is 90 m .