Question

Question #4e623

Answer 1

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`

Answer 2

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.

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

Answer 3

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.