Question #4e623

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#

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#

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.