Question

A speedboat starts from rest with a constant acceleration of +1.50 m/s2. After 4.00 seconds, its acceleration changes to +1.00 m/s2. After another 5.00 seconds, its acceleration changes to zero and the boat travels with constant velocity for a further 10.0 seconds. The boat then slows down and stops by accelerating at -2.00 m/s2.

1. Calculate the velocity and displacement of the boat at t = 4.00 seconds.

2. Calculate the velocity and displacement of the boat at t = 9.00 seconds

3. Calculate the time taken for the boat to slow down from its maximum speed to rest.

4. Calculate the total displacement vector of the boat in its journey.

Answer 1

Initial velocity of the boat = V₀ = 0 m/s (The boat starts from rest)

t₀ = 0 sec

t₁ = 4 sec

t₂ = 9 sec

t₃ = 19 sec

t₄ = Time at which the boat comes to a stop

Acceleration of the boat from t=0 to t=4 sec = a₁ = 1.5 m/s²

Velocity of the boat at t=4sec = V₁

Displacement of the boat from t=0 to t=4 sec = s₁

V₁ = V₀ + a₁(t₁ - t₀)

V₁ = 0 + (1.5)(4 - 0)

V₁ = 6 m/s

s₁ = V₀(t₁ - t₀) + a₁(t₁ - t₀)²/2

s₁ = (0)(4 - 0) + (1.5)(4 - 0)²/2

s₁ = 12 m

Acceleration of the boat from t=4 to t=9 sec = a₂ = 1 m/s²

Velocity of the boat at t=9sec = V₂

Displacement of the boat from t=4 to t=9 sec = s₂

V₂ = V₁ + a₂(t₂ - t₁)

V₂ = 6 + (1)(9 - 4)

V₂ = 11 m/s

s₂ = V₁(t₂ - t₁) + a₂(t₂ - t₁)²/2

s₂ = (6)(9 - 4) + (1)(9 - 4)²/2

s₂ = 42.5 m

Acceleration of the boat from t=9 to t=19 sec = a₃ = 0 m/s² (The boat travels at constant velocity)

Velocity of the boat at t=19sec = V₃ = V₂ = 11 m/s (The boat travels at constant velocity)

Displacement of the boat from t=9 to t=19 sec = s₃

s₃ = V₂(t₃ - t₂) + a₃(t₃ - t₂)²/2

s₃ = (11)(19 - 9) + (0)(19 - 9)²/2

s₃ = 110 m

Acceleration of the boat from t=19 to t=t₄ sec = a₄ = -2 m/s²

Velocity of the boat at t=t₄sec = V₄ = 0 m/s (the boat comes to stop)

Displacement of the boat from t=19 to t=t₄ sec = s₄

V₄ = V₃ + a₄(t₄ - t₃)

0 = 11 + (-2)(t₄ - 19)

t₄ = 24.5 sec

s₄ = V₃(t₄ - t₃) + a₄(t₄ - t₃)²/2

s₄ = (11)(24.5 - 19) + (-2)(24.5 - 19)²/2

s₄ = 30.25 m

Time taken by the boat to slow down from its maximum speed to rest = t₅

t₅ = t₄ - t₃

t₅ = 24.5 - 19

t₅ = 5.5 sec

Displacement of the boat t=9sec = s₅

s₅ = s₁ + s₂

s₅ = 12 + 42.5

s₅ = 54.5 m

Total displacement of the boat = s

s = s₁ + s₂ + s₃ + s₄

s = 12 + 42.5 + 110 + 30.25

s = 194.75 m

1) Velocity of the boat at t=4sec = 6 m/s

Displacement of the boat at t=4sec = 12 m

2) Velocity of the boat at t=9sec = 11 m/s

Displacement of the boat at t=9sec = 54.5 m

3) Time taken by the boat to slow down from its maximum speed to rest = 5.5 sec

4) Total displacement of the boat in the journey = 194.75 m