Question

Difficulty: MediumSpeed, Velocity, and Acceleration

Calculate the acceleration for each of the following pieces of equipment. Arrange the scenarios in order from the greatest acceleration to the least acceleration.

  1. 1An emergency response boat increasing its speed from 5.0 m/s5.0 \text{ m/s} to 25.0 m/s25.0 \text{ m/s} over 8.0 seconds8.0 \text{ seconds}.
  2. 2A heavy towing winch pulling a disabled vehicle from 1.0 m/s1.0 \text{ m/s} to 4.0 m/s4.0 \text{ m/s} in 2.0 seconds2.0 \text{ seconds}.
  3. 3A runway snowplow accelerating from a complete stop to 12.0 m/s12.0 \text{ m/s} in 10.0 seconds10.0 \text{ seconds}.
  4. 4A construction crane lifting a concrete barrier from rest to 3.0 m/s3.0 \text{ m/s} in 6.0 seconds6.0 \text{ seconds}.

Answer

The correct order from greatest to least acceleration is: Emergency response boat (2.5 m/s22.5 \text{ m/s}^2), Heavy towing winch (1.5 m/s21.5 \text{ m/s}^2), Runway snowplow (1.2 m/s21.2 \text{ m/s}^2), and Construction crane (0.5 m/s20.5 \text{ m/s}^2).
To determine the correct order, we must calculate the magnitude of acceleration for each scenario using the formula a=vf−vita = \frac{v_f - v_i}{t}. The boat has the highest acceleration (2.5 m/s22.5 \text{ m/s}^2), followed by the winch (1.5 m/s21.5 \text{ m/s}^2), the snowplow (1.2 m/s21.2 \text{ m/s}^2), and finally the crane (0.5 m/s20.5 \text{ m/s}^2). Arranging them from greatest to least yields this precise sequence.

Step-by-Step Solution

1
Calculate the acceleration of the emergency response boat.
The boat accelerates at 2.5 m/s22.5 \text{ m/s}^2.
Using the formula a=vf−vita = \frac{v_f - v_i}{t}, we get a=25.0−5.08.0=2.5 m/s2a = \frac{25.0 - 5.0}{8.0} = 2.5 \text{ m/s}^2.
2
Calculate the acceleration of the heavy towing winch.
The winch accelerates at 1.5 m/s21.5 \text{ m/s}^2.
Using the formula a=vf−vita = \frac{v_f - v_i}{t}, we get a=4.0−1.02.0=1.5 m/s2a = \frac{4.0 - 1.0}{2.0} = 1.5 \text{ m/s}^2.
3
Calculate the acceleration of the runway snowplow.
The snowplow accelerates at 1.2 m/s21.2 \text{ m/s}^2.
From a complete stop, vi=0v_i = 0. a=12.0−010.0=1.2 m/s2a = \frac{12.0 - 0}{10.0} = 1.2 \text{ m/s}^2.
4
Calculate the acceleration of the construction crane.
The crane accelerates at 0.5 m/s20.5 \text{ m/s}^2.
From rest, vi=0v_i = 0. a=3.0−06.0=0.5 m/s2a = \frac{3.0 - 0}{6.0} = 0.5 \text{ m/s}^2.
5
Order the calculated acceleration values from greatest to least.
Boat (2.52.5) > Winch (1.51.5) > Snowplow (1.21.2) > Crane (0.50.5).
The prompt requires arranging the scenarios in descending order based on their calculated magnitudes of acceleration.

Key Concept

Acceleration is defined as the rate at which an object changes its velocity over a specific interval of time.
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