You are driving your car and the traffic light ahead turns red. You applied the brake for 3.00 seconds and the velocity of the car decreases to 4.50 m/s. The car's deceleration has a magnitude of 2.70 m/s^2 during this time. What is the car's displacement?

this is too advanced for me bro

V = Vo + a*t = 4.50 m/s.

Vo + (-2.7)*3 = 4.5
Vo = 4.5 + 8.1 = 12.6 m/s.

D = Vo*t + 0.5a*t^2
D = 12.6*3 - 0.5*2.7*3^2 = 25.7 m.

Oh, I see we're getting into some physics here. Alright, buckle up, and let's calculate the car's displacement!

To start, we need to use the equation of motion:

vf = vi + at

Here, vf represents the final velocity (4.50 m/s) and vi is the initial velocity (which we assume to be zero since the car comes to a stop). The acceleration (a) is given as -2.70 m/s^2 (negative because deceleration).

Plugging in the values, we get:

4.50 m/s = 0 + (-2.70 m/s^2)t

Simplifying, we find:

t = 4.50 m/s / 2.70 m/s^2 ≈ 1.67 s

Now that we know the time taken for the car to decelerate, we can calculate the displacement using the equation:

d = vit + 1/2 * at^2

Since the initial velocity is zero, the equation simplifies to:

d = 1/2 * at^2

Plugging in the values, we get:

d = 1/2 * (-2.70 m/s^2) * (1.67 s)^2

Calculating that out, we find:

d ≈ -3.55 m

Now, before you get worried about the negative sign, remember that it just indicates the direction of motion. In this case, the negative sign means that the displacement is in the opposite direction of the car's initial motion.

So, the car's displacement is approximately -3.55 meters. I hope I didn't drive you crazy with all these calculations!

To find the car's displacement, we can use the kinematic equation:

\(v^2 = u^2 + 2as\)

Where:
- \(v\) is the final velocity (4.50 m/s)
- \(u\) is the initial velocity (which we need to find)
- \(a\) is the acceleration (negative value since it is deceleration, -2.70 m/s^2)
- \(s\) is the displacement (unknown)

First, let's find the initial velocity (\(u\)) using the given information.

\(v^2 = u^2 + 2as\)

Rearranging this equation to solve for \(u\), we get:

\(u^2 = v^2 - 2as\)

Substituting the values:

\(u^2 = (4.50 \, \text{m/s})^2 - 2 \times (-2.70 \, \text{m/s}^2) \times s\)

Simplifying further:

\(u^2 = 20.25 \, \text{m}^2/\text{s}^2 + 5.4 \, \text{m/s}^2 \times s\)

To solve for \(u^2\) and eliminate \(u\), we need another equation that relates displacement \(s\) and time \(t\).

We can use the equation of motion:

\(v = u + at\)

Where:
- \(v\) is the final velocity (4.50 m/s)
- \(u\) is the initial velocity (unknown)
- \(a\) is the acceleration (given as -2.70 m/s^2)
- \(t\) is the time (3.00 seconds)

Rearranging the equation to solve for \(u\), we get:

\(u = v - at\)

Substituting the values:

\(u = 4.50 \, \text{m/s} - (-2.70 \, \text{m/s}^2) \times 3.00 \, \text{s}\)

\(u = 4.50 \, \text{m/s} + 8.10 \, \text{m/s}\)

\(u = 12.6 \, \text{m/s}\)

Now that we have the value for \(u\), we can substitute it back into the previous equation to solve for \(s\):

\(u^2 = v^2 - 2as\)

\((12.6 \, \text{m/s})^2 = (4.50 \, \text{m/s})^2 - 2 \times (-2.70 \, \text{m/s}^2) \times s\)

\(158.76 \, \text{m}^2/\text{s}^2 = 20.25 \, \text{m}^2/\text{s}^2 + 5.4 \, \text{m/s}^2 \times s\)

Rearranging further:

\(5.4 \, \text{m/s}^2 \times s = 158.76 \, \text{m}^2/\text{s}^2 - 20.25 \, \text{m}^2/\text{s}^2\)

\(5.4 \, \text{m/s}^2 \times s = 138.51 \, \text{m}^2/\text{s}^2\)

Dividing both sides by \(5.4 \, \text{m/s}^2\), we get:

\(s = 138.51 \, \text{m}^2/\text{s}^2 \div 5.4 \, \text{m/s}^2\)

\(s \approx 25.65 \, \text{m}\)

Therefore, the car's displacement is approximately 25.65 meters.

25.65 m