The correct question is:
A motorboat starts from rest (initial velocity v(0) = v_0 = 0). Its motor provides a constant acceleration of 4 ft/s², but water resistance causes a deceleration of v²/400 ft/s².
(a) Write a differential equation for the velocity v.
(b) Find v when t = 10s, and also find the limiting velocity as t → ∞ (that is, the maximum possible speed of the boat).
Answer:
(a) The differential equation is:
dv/dt = 4 - (v²/400)
(b) v = 30.46ft/s when t = 10s.
(c) The limiting velocity as t → ∞ is 40ft/s
Step-by-step explanation:
(a) The differential equation is formed by considering the changes in acceleration. At a constant acceleration of 4 ft/s², it then decelerate at v²/400 ft/s².
Adding these, we have
Acceleration, a = 4 - (v²/400)
Because acceleration is the derivative of velocity, a = dv/dt.
dv/dt = 4 - (v²/400)
(b) The differential equation obtained in (a) is solved, and evaluated at t = 10, to obtain v = 30.46 ft/s
(c) The limiting velocity, taking into consideration the properties of limit is evaluated to obtain v = 40 ft/s.
SEE ATTACHMENT FOR THE STEP BY STEP WORKINGS.
