Question

As you approach the dock, you cut the engine of your motorboat and allow it to coast toward the dock. The speed of the boat during this time is given by v = vi^e−kt where vi is the speed of the boat at t = 0, v is its speed at time t, and k is a constant. The speed at the moment you cut the engine is 13.5 m/s and its speed at t = 17.8 s is 6.80 m/s. (Assume that the initial velocity of the boat is pointing in the positive direction.)

(a) How fast are you moving at t = 32.0 s?

(b) Determine the equation that describes the acceleration of the boat. (Use the following as necessary: k, v. Indicate the direction with the sign of your answer.)

(c) What is the acceleration of the boat at t = 32.0 s? (Indicate the direction with the sign of your answer.)

Answer 1

.

given, v = vi*e^(-k*t)

at t = 0 , vi = 13.5 m/s

at t = 17.8 s, v = 6.80 m/s

lets find k by using this relations.

v = vi*e^(-k*t)

6.80 = 13.5*e^(-k*17.8)

6.8/13.5 = e^(-k*17.8)

ln(6.8/13.5) = -k*17.8

==> k = -ln(6.8/13.5)/17.8

= 0.03853 s^-1

a) at t = 32.0 s

v = vi*e^(-k*t)

= 13.5*e^(-0.03853*32)

= 3.93 m/s <<<<<<<<<<<<<<<---------------------Answer

b) a = -dv/dt

= -d(vi*e^)(-k*t))/dt

= vi*e^(-k*t)*(-k)

= -vi*k*e^(-k*t) <<<<<<<<<<<<<<<---------------------Answer

c) at t = 32.0 s

a = -13.5*0.03853*e^(-0.03853*32)

= -0.152 m/s^2 <<<<<<<<<<<<<<<---------------------Answer