Finding Acceleration: Solving a Simple Velocity Equation at 0.5s

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Homework Statement


Velocity in m/s at time t is defined as v= 20t(1+2t)^-2

Find Acceleration at 0.5s

2. The attempt at a solution

I missed the day we covered this topic in class and I am under the assumption that I would find the derivative of 20t(1+2t)^-2 and then find the derivative once more to get acceleration. Once I get the second derivative its as easy as substituting 0.5 for t. Unfortunately I never get the correct answer which I was told is 9.6m/s. Is the answer wrong? I end up with 15m/s all the time.
 
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If you are given the velocity function you only need to differentiate once to get the acceleration.
 
I must be doing something wrong I keep getting the wrong answers.

Example:

s(t) = (4t^2 + 5)^3 t=1

I do the work

First Derivative - 24t(4t^2 + 5)^2
Second Derivative - 384t^2(4t^2 + 5)

My answer is 3456 m/s and the book tells me it is 5400 m/s
 
So is this supposed to be a new question? What happened with the first one you posted?

Note the unit of acceleration is m/s^2.

Your second derivative is not right. You're using the product rule, right? You're missing a term.
 
:smile: I feel really stupid right now. I knew something was up, turns out I was forgetting the product rule for all the questions. Thanks for clueing me into that. What a silly mistake.
 
You're welcome, lol. Those things happen. :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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