A car stops for a red ligth. The light turns green and the car moves for 3 seconds at a steadily increasing speed. During this time, it travels 20 meters. The car then travels at a constant speed for another 3 seconds for a distance of 30 meters. Finally, when approaching another red light, the car steadily slows to a stop during the next 3 seconds in 15 meters. Create equations with respect to time for the position and velocity of the car at each segment.

Question

A car stops for a red ligth. The light turns green and the car moves for 3 seconds at a steadily increasing speed. During this time, it travels 20 meters. The car then travels at a constant speed for another 3 seconds for a distance of 30 meters. Finally, when approaching another red light, the car steadily slows to a stop during the next 3 seconds in 15 meters. Create equations with respect to time for the position and velocity of the car at each segment.

Answer 1

To create equations for the position and velocity of the car at each segment, we can make use of the kinematic equations of motion.

Segment 1: Car moving from rest for 3 seconds.

Let's assume the initial position of the car (at t = 0) is x = 0, and its initial velocity (at t = 0) is v = 0.

Equation of motion for position:
x1(t) = 1/2 * a * t^2 + v0 * t + x0

Since the car is initially at rest (v0 = 0), the equation simplifies to:
x1(t) = 1/2 * a * t^2

Equation of motion for velocity:
v1(t) = a * t + v0

Again, since v0 = 0, the equation simplifies to:
v1(t) = a * t

Segment 2: Car moving at a constant speed for 3 seconds.

During this segment, the car moves with a constant velocity, so its acceleration (a) is 0.

Equation of motion for position:
x2(t) = v2 * t + x1(3)

Since the car traveled 20 meters during the first segment, we can substitute x1(3) = 20 to get:
x2(t) = v2 * t + 20

Equation of motion for velocity:
v2(t) = v2

Segment 3: Car decelerating to a stop for 3 seconds.

The car decelerates uniformly, so its acceleration (a) is negative.

Equation of motion for position:
x3(t) = -1/2 * a * (t - 3)^2 + v2 * (t - 3) + x2(3)

Since the car traveled 30 meters during the second segment, we can substitute x2(3) = 30 to get:
x3(t) = -1/2 * a * (t - 3)^2 + v2 * (t - 3) + 30

Equation of motion for velocity:
v3(t) = -a * (t - 3) + v2

These equations represent the position (x) and velocity (v) of the car at different times (t) during each segment.

Answer 2

d1 = 0.5a*t^2 = 20 m.

0.5a*3^2 = 20,
4.5a = 20,
a = 4.44 m/s^2.
V = at = 4.44 m/s^2 * 3 s=13.32 m/s.

V = d2/t = 30/3 = 10 m/s.
a = (Vf-Vo)/t = (0-10) / 3=-3.33 m/s^2

d3 = Vo*t + 0.5a*t^2 = 15 m.
10*3 + 0.5a*3^2 = 15,
30 + 4.5a = 15
4.5a = 15-30 = -15,
a = --3.33 m/s^2.