Question

The velocity of a particle is v = v0+gt+ft^2. If its position is x = 0 at t=0, then its displacement after unit time (t=1 s) is

Answer 1

this is the answer for your question

Answer 2

Answer:

The displacement after unit time is $v_{0}+\dfrac{g}{2}+\dfrac{f}{3}$.

Explanation:

The velocity is given by

$v = v_{0}+gt+ft^2$

We know that,

The velocity is

$v = \dfrac{dx}{dt}$

$dx=v dt$

If its position is x = 0 at t=0,

On integrating both side

$\int_{0}^{x}dx=\int_{0}^{1}(v_{0}+gt+ft^2)dt$

$[x]_{0}^{x}=[v_{0}+g\dfrac{t^2}{2}+f\dfrac{t^3}{3}]_{0}^{1}$

$x=v_{0}+\dfrac{g}{2}+\dfrac{f}{3}$

Hence, The displacement after unit time is $v_{0}+\dfrac{g}{2}+\dfrac{f}{3}$.