Question

the displacement x of a particle along x-axis is given by X equal to t square + 4t + 8 find the initial velocity of the particle​

Answer 1

Answer:

4

Explanation:

X=t^2+4t+8

For velocity d(x) /dt=d(t^2+4t+8)/dt

d(v) =2t+4

For initial velocity we take t=0

Initial velocity =4

Answer 2

Answer:

4

x =2plus 4&1x=4 now follow me