# Velocity at x seconds, given initial velocity and initial position

The equation for acceleration as a function of time is $a(t)=ct$ where $c= 3.71 \,\mathrm m\mathrm s ^ {-3}$. If the initial velocity is $v_0 = 3.24\, \mathrm m\mathrm s$ and the initial position is $x_0 = 1.35\,\mathrm m$, what is the velocity at $t_1=2.00 \,\mathrm s$?

I'm given the acceleration formula. To get the velocity, I take its integral, correct? That would be $\dfrac{ct^2}{2} + C$. I plug in $2.00\,\mathrm s$ for $t$ and get $(9.92 + C) \,\mathrm{m} \mathrm{s}^{-1}$.

I was told this is wrong and the correct answer is $13.2\, \mathrm{m} \mathrm{s}^{-1}$. How do you arrive at this solution?

## 1 Answer

I excluded the step of plugging in the initial velocity as the constant. After plugging it in, I get

$$9.92 \,\frac{\mathrm m}{\mathrm s}+ 3.24 \,\frac{\mathrm m}{\mathrm s} = 13.16 \,\frac{\mathrm m}{\mathrm s} \approx \boxed{13.2 \,\frac{\mathrm m}{\mathrm s}}$$

credit: erenust