What value of t does the acceleration equal 0? If time has to be between 0 and 4, what does that imply about t?
Also if you have multiple solutions, which ones are maximums (hint: second derivative tests)?
Given that the motion of an object is given by v(t) = , where t is in seconds. Find the maximum velocity of the object for the first four seconds.
I know maximum velocity would mean acceleration =0. So I differentiated v(t), and equated it to 0, but I do not really know what to do next..
Thank you for your help!
Think about what the graphs will look like: what is the relationship between velocity and acceleration? If you are looking for a maximum velocity across the domain [0, 4] seconds, then you are looking for the highest value that will have (that is, the maximum value it reaches on the axis. Rolle's Theorem states that, for a continuous, differentiable function (in your case , across the open interval (those two red dots), there is some value (blue dot) for which the instantaneous slope is zero. It could be a maximum or a minimum. That is what you will have to find out.
So, we know that is our velocity, if we take the first derivative, it will tell us where the zero-points are (those maximums or minimums) because the instantaneous slope at those points are zero. Try to imagine putting a ruler next to the upside down parabola; as you swing it from -4 to 0, the slope starts to decrease. At 0, you can see that is where the ruler would be completely horizontal (meaning it has a slope of zero).
Dividing both sides by -1 would remove the negative sign (since it is zero):
Now this is where you come in! Over what intervals from [0, 4] for the function will the result be zero? Those are your possible minimums or maximums.
Once you obtain your possible maximums or minimums, you need to use the 2nd derivative test to distinguish between them. Recall the following:
Second Derivative Test
- y'' is positive → Curve is concave up (Minimum)
- y'' is negative → Curve is concave down (Maximum)
- y'' is zero → Possible inflection point (where concavity changes)
So you calculate the second derivative, enter them into , and see if the results are positive or negative. That will tell you where your maximum velocity is (note: there could be multiple, especially with a periodic function).
Hi, thank you both! I'm not exactly sure still though,
When I put v'(t) to 0, I'd get pi/4
When I put pi/4 into v(t), I get 0.
So I differentiated again to get v''(t), and I got ..I put subbed t=pi/4 here to see whether it is a negative or not, and I got a 0..
Nevertheless, thank you!
Then you'll see that the maximum velocity FOR THE FIRST FOUR SECONDS occurs when t=0. And that, if for example, the domain had been time going from 0 to 7 seconds , THEN, your maximum velocity would have occurred at t=7pi/4 seconds = 5.5 seconds.
So here, it's not a case of blindly applying at maximum velocity, dv/dt = 0. You're dealing with a periodic function.
So, I'll advise going back to learning how to draw graphs of trigonometrical functions, getting re-acquainted with amplitude, period, etc...