Question

A particle is moving along a straight line from P towards Q. When it crosses the point A, its velocity is 10 m.s-1. After 2 s, it crosses another point B with velocity 6 m.s-1. If the particle moves with uniform acceleration between A and B, find the magnitude and direction of this acceleration. Pls answers fast!!!°°°

Answer 1

Answer:

Acceleration between A and B

                            = (6 – 10) m/s ÷ (2-0) s  

                             = -4 m/s ÷ 2 s

                            = -2 m/s2

∴ Magnitude of between A and B acceleration between A and B = 2m/s2

∴ Direction of the acceleration is opposite to that of the velocity.

HOPE IT HELPS YOU!!!

PS : I am from SPHS also....!!

Answer 2

It is given that the acceleration of the particle is uniform.

Velocity when the particle crosses point A, initial velocity, u=10ms-1.

Velocity when the particle crosses point B, final velocity, v= 6ms-1.

Time between A and B, t=2s

∴v=u+at

=>6=10+a×2

=> 2a=-4

=> a=(-4)/2=-2ms^(-2)

The magnitude of the acceleration is -2ms^-2 or retardation is 2ms^-2 and the direction of the acceleration is opposite to that of the initial velocity. (ANS)

Hope it helps.