Answer to Question #79637 in Physics for Anuj

Question

A particle moves along X direction with constant acceleration.Its velocity after 4s is 18m/s and displacement is 56 m.Find out its initial velocity and acceleration.

Ans is 10m/s,2m/s^2 but method don't know

Answer 1

a) We can find the initial velocity of the particle from the kinematic equation:
d=(v_0+v)/2∙t,
here, v_0 is the initial velocity of the particle, v=18 m⁄s is the final velocity of the particle after time t=4 s and d=56 m is the displacement of the particle.
Then, we get:
v_0=2d/t-v=(2∙56 m)/(4 s)-18 m/s=10 m/s.
b) We can find acceleration of the particle from another kinematic equation:
v=v_0+at,
a=(v-v_0)/t=(18 m/s-10 m/s)/(4 s)=2 m/s^2 .