Stationary Hovercraft

Question

An initially stationary hovercraft sits on a large lake. When a whistle blows, the hovercraft accelerates due north at a rate of 1.2m s^-2 for 10s, does not accelerate at all for the next 10s, and then accelerates at a rate of 0.6m s^-2 due east for another 10s. The hovercraft then coasts for another 10s without any acceleration.

a) What is the velocity of the hovercraft 40s after the whistle blows?

b) What is the displacement of the hovercraft 40s after the whistle blows?

Answer 1

Hi Stephany B.,

Well use equations v = v₀ + at and d = d₀ + v₀t + 1/2at²

For the first 10 seconds the hovercraft accelerates and the velocity and distance are:

v_N = 1.2 m/s² *10 s = 12 m/s, (v₀ = 0)

d_N = 1/2*1.2 m/s² * 10² s² = 60 m

For 10 to 20 seconds:

v_N = 12 m/s, (no change)

d_N = 12 m/s*10 s = 120 m

For 20 to 30 seconds:

v_N = 12 m/s, (no change, except we now have velocity in two directions, North and East)

d_N = 12 m/s*10 s = 120 m

v_E = 0.6 m/s² *10 s = 6 m/s

d_E = 1/2*0.6 m/s²*10² s² = 30 m

For 30 to 40 seconds:

v_N = 12 m/s

d_N = 12 m/s*10 s = 120 m

v_E = 6 m/s

d_E = 6 m/s*10 s = 60 m

Final velocity is 12 m/s North and 6 m/s East:

[ 12² + 6² ]^1/2 = 13.4 m/s

tan^-1(12/6) = 63.4^o NE

Velocity after 40 seconds is v_f = 13.4 m/s 63.4^o NE

Final total displacement is (displacement is final position minus initial position, not equal to total distance):

d_N = (60 + 120 + 120 + 120) m = 420 m

d_E = (30 + 60) m = 90 m

d_d = [420² + 90²]^1/2 = 429.5 m displacement after 40 seconds.

I hope this helps, Joe.