A rapid transit commuter train leaves a station, accelerating uniformly along a straight t… — Armed Services Vocational Aptitude Battery (ASVAB)

A rapid transit commuter train leaves a station, accelerating uniformly along a straight track. It reaches a cruising velocity of $27.0 \text{ m/s}$ from a complete stop. If the acceleration phase lasts exactly $0.75 \text{ minutes}$ , what is the train's average acceleration in $\text{m/s}^2$ ?

Answer: 0.6 m/s^2

Answer

The train's average acceleration is

0.6 \text{ m/s}^2

Acceleration is defined as the change in velocity divided by the time it takes for that change to occur. By converting the time unit to seconds and substituting the initial velocity, final velocity, and time into the standard acceleration formula, the correct magnitude of

0.6 \text{ m/s}^2

is determined.

Step-by-Step Solution

Identify the known kinematic variables.

Initial velocity

v_i = 0 \text{ m/s}

, final velocity

v_f = 27.0 \text{ m/s}

, and time

t = 0.75 \text{ minutes}

Extracting the given values is necessary before selecting the appropriate formula.

Convert the time from minutes to standard SI units (seconds).

t = 0.75 \times 60 = 45 \text{ seconds}

Acceleration in

\text{m/s}^2

requires the time unit to match the time component of the velocity unit (seconds).

Calculate the average acceleration using the formula

a = \frac{v_f - v_i}{t}

a = \frac{27.0 - 0}{45} = 0.6 \text{ m/s}^2

This formula directly relates the change in an object's velocity to the time interval over which the change occurred.

Key Concept

Kinematics: Uniform Acceleration and Unit Conversion

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic