Question #8ecac

1 Answer
Oct 11, 2017

Well, let's takes #f'(t)=10e^(1.05)t=10e^(20/21t)#

If #y=ae^(bx)#, where #a# and #b# are constants, then #(dy)/(dx)=abe^(bx)#.

So, #10e^(1.05t)=abe^(bt)#.

Therefore, #a*1.05=10#.

#a=10/1.05=200/21#

So, #intf(v) dt=(200e^(1.05t))/21#

This will be rewritten as #v_y=(200e^(1.05t))/21#

This will now be rearranged to get #t=ln((21v_y)/200)/1.05#

To find the total time taken, you do #ln((21v_2)/200)/1.05 - ln((21v_1)/200)/1.05#. Where #v_2# is the final velocity, and #v_1# is the initial velocity.

This is all I can give you since I don't have the full question, only what you gave which was a part 4 to a, and b.

If your value for #t# is negative, just make it positive by taking the value of #abs(t)#