# Using definite integrals with velocity and acceleration

Rocket A is traveling 49 ft/sec at 80 seconds. Rocket B is launched upward with an acceleration of $$a(t)=\frac3{\sqrt{t +1}}$$. At time t=0 seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at t=80 seconds?

• What is acceleration compared to velocity ? – Claude Leibovici Feb 26 '14 at 17:02
• Acceleration is the derivative of velocity – Hannah Feb 26 '14 at 17:03
• OK. So, velocity is the ??? of acceleration, isn't ? – Claude Leibovici Feb 26 '14 at 17:04
• yes I understand the relationship between acceleration and velocity however I don't understand how to use the information given to solve using the relationship. – Hannah Feb 26 '14 at 17:15
• So, velocity is the antiderivative of acceleration. Is this way of explaining making things clearer ? We have the same with velocity and distance. Let us continue until you are sure. Getting the answer is not important; understanding is ! – Claude Leibovici Feb 26 '14 at 17:32

$a(t) = \frac{dv}{dt}$ where v(t) is the velocity.
Then $v(t) = \int{a(t)dt}$. Use the condition for velocity at $t = 0$ to obtain the integration constant. Then you can get velocity at $t = 80$.
• You do the integral shown. You will come out with $v(t)=something+C$, where $something$ is the result of the integral. Now substitute in $t=0, v(0)=2$ and you will get an equation in $C$ – Ross Millikan Feb 26 '14 at 17:22
• $c$ can be anything based on the condition. Why do you think you should get $c = 0$? – R.K. Feb 26 '14 at 17:41