A coastal patrol boat is cruising at an initial velocity of 5.0 \text{ m/s}. The captain o… — Armed Services Vocational Aptitude Battery (ASVAB)

A coastal patrol boat is cruising at an initial velocity of $5.0 \text{ m/s}$ . The captain orders the engines to increase power, causing the boat to accelerate at a constant rate of $0.2 \text{ m/s}^2$ for exactly $1.5 \text{ minutes}$ .

What is the final velocity of the boat in $\text{m/s}$ ?

Answer: 23 m/s

Answer

23.0

The correct answer is derived by first converting the time from minutes to seconds (

1.5 \text{ min} = 90 \text{ s}

), calculating the total change in velocity (

0.2 \times 90 = 18.0 \text{ m/s}

), and then adding that change to the initial cruising velocity (

5.0 + 18.0 = 23.0 \text{ m/s}

Step-by-Step Solution

Extract and convert the time variable into seconds.

t = 1.5 \text{ minutes} \times 60 \text{ seconds/minute} = 90 \text{ seconds}

Acceleration is given in meters per second squared (

\text{m/s}^2

), so time must be in seconds to ensure unit consistency.

Calculate the total change in velocity during the acceleration phase.

\Delta v = a \times t = 0.2 \text{ m/s}^2 \times 90 \text{ s} = 18.0 \text{ m/s}

The change in velocity is the product of the constant acceleration and the duration of that acceleration.

Add the change in velocity to the initial velocity.

v_f = v_i + \Delta v = 5.0 \text{ m/s} + 18.0 \text{ m/s} = 23.0 \text{ m/s}

The final velocity is the starting velocity plus the amount the velocity increased.

Key Concept

Applying the constant acceleration formula (

v_f = v_i + at

) with necessary unit conversions.

Estimated Time:1m 0s

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