Question #4e623

3 Answers
May 26, 2017

You find the time of maximum velocity by taking the derivative of the velocity function and setting it equal to 0. Check if the value(s) is/are (a) minimum(s) or maximum(s), and if it is a maximum, plug in the time into the velocity function to get your answer. The maximum velocity of that function is 64 m/s.

Explanation:

So I will explain your error in detail first. What you found by setting dx/dt = 0 was a critical point for x(t), the position function. This means that there is a turning point at t=16 for the function (I think you meant to divide instead of multiply in step 3 but it would have gotten you to the same answer), which indicates a maximum or minimum of the position function.

So to solve your problem, you find the derivative of the velocity function to find your acceleration function (d^2x)/(dt^2).

The derivative of t(16-t), the velocity function, is given by the product rule (1*d2 +2*d1). Using this we get -t + (16-t), or -2t + 16 as (d^2x)/(dt^2).

We set this acceleration function to 0 so that 0= -2t+16. Subtract 16 from both sides to get -16=-2t. Divide by -2 to get t=8.

This is the time where a minimum or maximum velocity is reached. We plug in values around this number to determine if it is a minimum or maximum. If values to the left have a negative slope while values to the right have a positive one, it is a minimum. It is vice versa for a maximum.

Plugging in t=7 into -2t + 16 gives 2 while plugging in 9 gives -2. This indicates a maximum at t=8.

Plug this value into your velocity function so that dx/dt = (8)(16-(8)) = 64m/s

Jul 28, 2017

There appears at typo in the question where it is stated that (dx)/dt=(16-t)
A case of missing t

Explanation:

Given expression for velocity is
(dx)/dt=t(16-t)

There are two methods for solving this question.

A. Graphical method. Draw the function in a graph and find out the maximum. I used built-in Graphical tool and the result is as follows. x axis is taken as time and y axis as velocity.
My computerMy computer

We see that maximum occurs as t=8s and maximum velocity (dx)/dt=64ms^-1

Jul 28, 2017

B. Using Calculus.

Explanation:

It can be rewritten as

v=16t-t^2

To find critical points of a function, we differentiate with respect to the variable and set it equal to 0.

(dv)/dt=d/dt(16t-t^2)
=>(dv)/dt=0=16-2t
=>2t=16
=>t=8s

Value of function for velocity at t=8 is

(dx)/dt|_(t=8)=16xx8-8^2

(dx)/dt|_(t=8)=128-64=64ms^-1

To check it for a maximum we need to calculate second differential of the function with respect to the variable and show that it has a -ve value at the point of interest.

(d^2v)/dt^2=d/dt(16-2t)
=>(d^2v)/dt^2=-2
It is -ve for all values of the variable. Hence a maximum.