A heavy transport truck is traveling East on a straight highway at a constant speed of 90 … — Armed Services Vocational Aptitude Battery (ASVAB)

A heavy transport truck is traveling East on a straight highway at a constant speed of $90 \text{ km/h}$ . The driver applies the brakes, producing a constant deceleration until the truck comes to a complete stop in exactly $10 \text{ seconds}$ . Which of the following correctly describes the truck's average acceleration vector during this braking period?

$2.5 \text{ m/s}^2$ WestAnswer
B
$2.5 \text{ m/s}^2$ East
C
$9.0 \text{ m/s}^2$ West
D
$9.0 \text{ m/s}^2$ East

Answer

The correct acceleration vector is

2.5 \text{ m/s}^2

West.

To find the acceleration, the initial speed must first be converted from km/h to m/s by multiplying by

1000/3600

, resulting in

25 \text{ m/s}

. The acceleration magnitude is the change in velocity (

25 \text{ m/s}

) divided by the time (

10 \text{ s}

), giving

2.5 \text{ m/s}^2

. Because the truck is braking while moving East, the acceleration vector must oppose the motion, meaning it points West.

Step-by-Step Solution

Convert the initial velocity from km/h to m/s.

90 \text{ km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = 25 \text{ m/s}

East.

Standard kinematic equations require consistent metric units, specifically meters and seconds, to yield an acceleration in

\text{m/s}^2

Calculate the magnitude of the average acceleration using the formula

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

a = \frac{0 \text{ m/s} - 25 \text{ m/s}}{10 \text{ s}} = -2.5 \text{ m/s}^2

The final velocity (

v_f

) is zero because the truck comes to a complete stop.

Determine the direction of the acceleration vector.

The acceleration is

2.5 \text{ m/s}^2

West.

Because the truck is slowing down, the acceleration vector must point in the opposite direction of its velocity (East). The negative sign in the calculation confirms this opposing direction.

Key Concept

Kinematic equations with unit conversions and the vector nature of acceleration

Estimated Time:1m 30s

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