91Ó°ÊÓ

In the study of kinematics, equations of motion are fundamental tools used to describe the relationships between displacement, velocity, acceleration, and time. These equations allow us to predict and understand an object's motion under constant acceleration. For example, to solve part (a) of the exercise, we used the first equation of motion:

• \( v_f = v_i + at \)

where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration (or deceleration, as in this case), and \( t \) is the time. This equation helped us determine the time it takes for the car to come to a stop.
To solve part (b), we used the second equation of motion:

• \( s = v_i t + \frac{1}{2} at^2 \)

This equation provided us with the total distance the car travels during the deceleration period. It considers both the initial velocity and the acceleration acting over time.

Deceleration is a specific type of acceleration where the velocity of an object decreases over time. This negative acceleration occurs in a direction opposite to the velocity, slowing down the object.
In the exercise, the car has a deceleration of \(-4.92 \ m/s^2\). This value indicates how quickly the car reduces its speed to zero. It's important to note that in physics, acceleration values are not always positive. For deceleration, the acceleration is negative, which is why the equation of motion uses this deceleration value to decrease the initial velocity.

A velocity-time graph is a visual representation that helps us understand an object's velocity as it changes over time. On this graph, time is plotted on the horizontal axis and velocity on the vertical axis.
In the exercise provided, the velocity-time graph is a straight line, sloping downward, reflecting the constant deceleration of the car:

• The starting point on the graph is at \( 24.6 \ m/s \), the initial velocity.
• The line descends steadily until it reaches \( 0 \ m/s \) at \( 5 \ s\), when the car stops.

This linear decrease shows the consistent reduction in speed over the 5 seconds it takes for the car to stop. The slope of the line corresponds to the deceleration rate of \(-4.92 \ m/s^2\).

A position-time graph illustrates an object's motion, showing how its position changes over time. Time is on the horizontal axis, while position is on the vertical axis.
In the problem, the position-time graph is a downward-opening parabola, indicating the car's decreasing speed:

• Initially, the car starts at position \(x = 0\) when \(t = 0\) seconds.
• As time progresses to \(t = 5\) seconds, it reaches a final position of \(61.5\) meters.

The parabolic shape signifies the influence of constant deceleration on position. Initially, the car moves rapidly, but as it decelerates, the rate of position change slows down, forming a curved graph. This result matches the equation \( s = v_i t + \frac{1}{2} at^2 \), highlighting the effect of both initial speed and deceleration.