A motorboat cuts its engine when its speed is 10.1 m/s and coasts to rest. The e

Question

A motorboat cuts its engine when its speed is 10.1 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 = 20.9 s, the speed is 5.00 m/s.

(a) Find the constant c.
____________s-1

(b) What is the speed at t = 40.0 s?
____________ m/s

(c) Differentiate the expression for v(t) and thus show that the acceleration of the boat is proportional to the speed at any time.

Explanation / Answer

a)

V = Vi*e^(-ct)

at t = 20.9 s, V = 5 m/s

Vi = 10.1

So, 5 = 10.1*e^(-c*10.1)

so, c = 0.0696 s-1 <------------answer

b)

speed at t = 40s , V = 10.1*e^(-0.0696*40) = 0.624 m/s

c)

a = dv/dt = d(Vi*e^(-ct))/dt

= Vi*(-c)*e^(-ct) = -c*(Vi*e^(-ct)) = -c*V <----a is proportional to V

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