Question

A motorboat cuts its engine when its speed is 8.2 m/s and coasts to rest. The equation describing the motion of the motorboat during this period is v = vie-ct, where v is the speed at time t, vi is the initial speed at t = 0, and c is a constant. At t = 18.4 s, the speed is 5.00 m/s.

(a) Find the constant c (in s-1)
(b) What is the speed at t = 40.0 s? (in m/s)

Answer 1

a.
`c  =  0.1739 (m)/(s^2)`

b.
`v(40.0 s ) = 1.64 (m)/(s)`

Step-by-step explanation:

So, the equation for the speed is:

`v(t) = v_i - c t`

as we know that the initial speed is

`v_i = 8.2 (m)/(s)`

and the speed at t=18.4 s is

`v(18.4 s ) = 8.2 (m)/(s)- c 18.4 s = 5.00 (m)/(s)`

Now, we can work it a little:

`- c 18.4 s = 5.00 (m)/(s) - 8.2 (m)/(s)`

`- c 18.4 s = -3.20 (m)/(s)`

`- c 18.4 s = -3.20 (m)/(s)`

`c  = ( -3.20 (m)/(s) )/( - 18.4 s )`

`c  =  0.1739 (m)/(s^2)`

So, at t=40.0 s the speed will be:

`v(40.0 s ) = 8.2 (m)/(s)- 0.1739 (m)/(s^2) * 40.0 s`

`v(40.0 s ) = 8.2 (m)/(s)- 6.96 (m)/(s)`

`v(40.0 s ) = 1.64 (m)/(s)`