In motion problems, understanding the relationship between acceleration, velocity, and speed is crucial. A common misconception is that positive acceleration always indicates an increase in speed. However, positive acceleration simply means that the velocity is becoming more positive, while negative acceleration indicates that the velocity is becoming more negative. This distinction is essential for correctly interpreting motion scenarios.

The formula for acceleration is given by:

$$ a = \frac{\Delta v}{\Delta t} $$

where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time. A positive value for acceleration indicates that the change in velocity is positive, while a negative value indicates a decrease in velocity.

To determine whether speed is increasing or decreasing, focus on the magnitude of the velocity, disregarding the sign. For instance, if an object moves from an initial velocity of 10 m/s to a final velocity of 30 m/s, the speed is increasing because the magnitude (30) is greater than the initial magnitude (10). Conversely, if an object moves from 30 m/s to 10 m/s, the speed is decreasing since the magnitude is reducing.

When analyzing acceleration and speed, it is important to note that:

• If the signs of acceleration and initial velocity are the same, the speed increases.
• If the signs of acceleration and initial velocity are different, the speed decreases.

For example, if an object has a positive initial velocity and positive acceleration, it speeds up. If it has a negative initial velocity and negative acceleration, it also speeds up. However, if the initial velocity is positive and the acceleration is negative, the object slows down, and similarly, if the initial velocity is negative and the acceleration is positive, the object also slows down.

To illustrate these concepts, consider a truck moving to the left at 30 m/s that comes to a stop after traveling 150 meters. The initial velocity is negative (since it moves left), and the final velocity is 0. Using the equation:

$$ v_f^2 = v_i^2 + 2a \Delta x $$

we can solve for acceleration. Substituting the known values, we find:

$$ 0 = (-30)^2 + 2a(-150) $$

which simplifies to:

$$ 0 = 900 - 300a $$

Rearranging gives:

$$ 300a = 900 $$

Thus, the acceleration is:

$$ a = 3 \, \text{m/s}^2 $$

Even though the truck is slowing down, the acceleration is positive because the direction of motion (left) is considered negative in this context.

Understanding these principles helps clarify the often confusing relationship between acceleration, velocity, and speed in motion problems.