$\begingroup$

A particle has zero velocity initially (i.e., at time $t=0$) and its acceleration at $t$ seconds is $a(t)=72t−4t^3$ meters-per-second per second. During the time interval $[5,8]$ seconds find the average acceleration and the average speed of the particle. I got the average acceleration by taking the integral and then plugging that into $f(8)-f(5)/8-5$ but I have tried working through the average speed and I can't find the answer for that one.

$\endgroup$ 1

3 Answers

$\begingroup$

Acceleration is the derivative of the velocity, $a(t) = \frac{dv}{dt}$, so you can find an expression for the velocity by integrating and then do the same averaging procedure that you did for the acceleration.

$\endgroup$ 1 $\begingroup$

Find the velocity at time $t$ by integrating $72t-4t^3$. We get $36t^2-t^4+C$. But the velocity at $t=0$ is $0$, and therefore $C=0$. So the velocity at time $t$ is given by $v(t)=36t^2-t^4$.

We are asked for the average speed from $t=5$ to $t=8$, not for the average velocity (trick question, some of us are nasty). Note that the velocity is negative when $t\gt 6$. So for the average speed, we find $$\int_5^6 (36t^2-t^4)\,dt+\int_6^8 (t^4-6t^2)\,dt,$$ and divide the result by the total elapsed time.

$\endgroup$ $\begingroup$

The integral of acceleration is velocity, and we are given a point defined as $v = 0$ when $t = 0$. We can then integrate acceleration and solve for c to get the equation of velocity.$$v(t)=\int{a(t)dt}$$ $$v(t)=\int{72t - 4t^4}$$ $$v(t) = 36t^2 - t^4 + c$$ $$0 = 0 + c$$ $$v(t) = 36t^2 - t^4$$Then, to find the average speed on the interval $[5, 8]$, find the velocity values for $t=5$ and $t=8$ and average those. $$v(5) = 275$$$$v(8) = -1,729$$The speed is just the absolute value of velocity, so the two speeds are $275$ and $1,729$.

Now, just average the two speeds. $$avg = \frac{275 + 1,729} {2}$$$$avg = 1,002 m/s$$

To find the average acceleration, simply follow the same procedure.$$a(5) = -140$$$$a(8) = -1472$$$$ avg = \frac{-140 - 1472}{2}$$$$ avg = -806 m/s^s$$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy