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?
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$\begingroup$ What is acceleration compared to velocity ? $\endgroup$– Claude LeiboviciCommented Feb 26, 2014 at 17:02
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$\begingroup$ Acceleration is the derivative of velocity $\endgroup$– HannahCommented Feb 26, 2014 at 17:03
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$\begingroup$ OK. So, velocity is the ??? of acceleration, isn't ? $\endgroup$– Claude LeiboviciCommented Feb 26, 2014 at 17:04
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$\begingroup$ 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. $\endgroup$– HannahCommented Feb 26, 2014 at 17:15
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$\begingroup$ 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 ! $\endgroup$– Claude LeiboviciCommented Feb 26, 2014 at 17:32
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1 Answer
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$a(t) = \frac{dv}{dt}$ 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 $t = 80$.
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$\begingroup$ How do you set up the finding the integration constant? $\endgroup$– HannahCommented Feb 26, 2014 at 17:08
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$\begingroup$ 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 $C$ $\endgroup$ Commented Feb 26, 2014 at 17:22
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$\begingroup$ So if i did it correctly i believe i should get c=0? $\endgroup$– HannahCommented Feb 26, 2014 at 17:37
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$\begingroup$ $c$ can be anything based on the condition. Why do you think you should get $c = 0$? $\endgroup$– R.K.Commented Feb 26, 2014 at 17:41
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$\begingroup$ I plugged in 2 for v(t) and 1 in for the t of my integrated equation and solved for c. So it looked like 2=2(0+1)^3/2 +c. I solved for c=0 $\endgroup$– HannahCommented Feb 26, 2014 at 17:50