A motorboat starts from rest (initial velocity v(0)=v0=0 ft/s. Its motor provides a constant acceleration of 4 ft/s2 but water resistance causes a deceleration of v2400 ft/s2. a) Write the differential equation for the velocity v. b) Find v when t=10s c) Find the limiting velocity as t→[infinity] (that is, the maximum possible speed of the boat.)

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).

(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.

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

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