Question

A particle travelling in a straight line with constant acceleration of 4ms passes at a point O when its velocity is 12ms-¹. It passes another point P after a further 3 seconds. Find the velocity of the particle at P and the distance OP.​

Answer 1

The acceleration by definition is the change in velocity. If the acceleration is constant at a rate of 4ms^-2, this means for every second traveled the particle gains 4m/s speed. In 3 seconds, it will gain 4ms^-2 x 3 = 12 m/s. If the starting velocity (Vo) is 12ms^-1, the new velocity at point P will be: 12 + 12 = 24 m/s.

The formula for distance (D) in an accelerating particle is: Vo x t + 0.5 x a x t^2. But if you substitute all terms, D =12 x 3 + 0.5 x 4 x 3^2
D = 54 m

Answer 2

See below

Step-by-step explanation:

Velocity calculation:

vf = vo + a t vf = final velocity vo = original velocity a = acceleration

= 12 + 4 ( 3) = 24 m/s

Displacement /Distance equation :

df = do + vo t + 1/2 a t^2

0 + 12 ( 3) + 1/2 ( 4) ( 9) = 54 m < === OP