Question

2. The acceleration of particle is a function of position is given by a(x) = 4 - 2x ms-2. Then velocity v(x) is equal to (given that v(0) = 0]​

Answer 1

Answer:

$\text{Velocity}\:\bf{V(x)=4x-x^2}$

Explanation:

Given:

Acceleration

$A=4-2x$

Integrating on both sides, we get

$\frac{dV}{dx}=4-2x$

$\implies\:dV=(4-2x)\:dx$

$\implies\:\int{dV}=\int{(4-2x)\:dx}$

$\implies\:V=4x-2\frac{x^2}{2}+c$

$\implies\:V(x)=4x-x^2+c$

But V(0)=0

$\implies\:4(0)-(0)^2+c=0$

$\implies\:c=0$

$\therefore\:\bf{V(x)=4x-x^2}$