Question

Question #8ecac

Answer 1

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)`