Question

Given the following velocity functions of an object moving along a line, find the position function with the given initial position v(t)=2t+4

Answer 1

Answer:

The position of the particle as a function of times is given by

$x(t)=t^{2}+2t+x(o)$  where $x(0)$ is the position of the particle at t = 0 secs.

Step-by-step explanation:

Give that

$v(t)=2t+4$

we know that

$v(t)=\frac{dx(t)}{dt}\\\\\therefore dx(t)=v(t)dt\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$x(t)=\int (2t+4)dt\\\\x(t)=\int 2tdt+\int 4dt\\\\x(t)=\frac{2t^{2}}{2}+4t+c$

here 'c' is the constant of integration which represents the position of the particle at time t = 0 secs.

Thus we have

$x(t)=t^{2}+2t+x(o)$