# A particle moves in a straight line so that at time t its displacement from a fixed origin is x and its velocity is v. If its acceleration is 4 + x and v = 1 when x = 0, what is v when x = 1?

May 22, 2018

The value of $v$ is $\frac{11}{2}$

#### Explanation:

Recall that the derivative of velocity is acceleration and the anti-derivative of acceleration is velocity.

$\int 4 + x \mathrm{dx} = \frac{1}{2} {x}^{2} + 4 x + C$

We can now solve for $C$.

$\frac{1}{2} {\left(0\right)}^{2} + 4 \left(0\right) + C = 1$

$C = 1$

Therefore the velocity function is $v = \frac{1}{2} {x}^{2} + 4 x + 1$ and the value of $v$ at $x = 1$ is hence $v \left(1\right) = \frac{1}{2} + 4 + 1 = \frac{11}{2}$

Hopefully this helps!