91Ó°ÊÓ

In this exercise, the concept of constant acceleration is pivotal. Acceleration refers to the rate at which an object changes its velocity. When acceleration is constant:

• the velocity of an object changes by the same amount each second;
• the object moves by increasingly larger distances in each second of motion.

In our scenario, you start from rest and accelerate towards the bus. Your acceleration is given as 0.960 m/s². This means each second, your velocity increases by 0.960 meters per second.
Given that your initial velocity (\(v_0\)) is 0 m/s, the formula for position under constant acceleration is: \(x_y(t) = \frac{1}{2} a t^2\). This allows us to calculate how far you have traveled over time. Constant acceleration helps predict the future state of motion, which is crucial for knowing how far you'll run to catch the bus.

The situation of catching the bus uses a quadratic equation, which is fundamental in kinematics. Quadratic equations are polynomials of degree two, typically written in the form: \(ax^2 + bx + c = 0\). They appear often in kinematics when dealing with uniformly accelerated motion.
In this problem, you must determine how long it takes you to reach the bus. This is done by setting your position, \(x_y(t) = \frac{1}{2} a t^2\), equal to the bus's position, \(x_b(t) = v_b t + d_0\), leading to \(0.480 t^2 - 5.00 t - 12.0 = 0\).
This re-arrangement forms a quadratic equation where:

• \(a = 0.480\), reflecting the half of acceleration;
• \(b = -5.00\), representing the bus's speed;
• \(c = -12.0\), the initial gap in positions.

The quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) solves this equation, providing the time it takes for the positions to equalize.

When determining your speed as you catch up with the bus, velocity calculation under constant acceleration is critical. Velocity gives the rate and direction of your movement. It's calculated via the formula: \(v = v_0 + at\).
In this context, with an initial velocity (\(v_0\)) of 0 m/s, your velocity when you catch up to the bus becomes: \(v = at\).
Substitute the acceleration (0.960 m/s²) and the time (\(t\)) from solving the quadratic equation into this formula, and you determine how fast you are moving when you reach the bus.
This step shows how an increase in time, given constant acceleration, directly influences your final speed, reinforcing why understanding velocity calculation is crucial for predicting motion outcomes in kinematics.