A red car and a green car, identical except for color, move toward each other in adjacent lanes and parallel to an x-axis. At time t=0, the red car is at x sub r = 0 and the green car is at x sub g = 220m. If the red car has a constant velocity of 20km/h, the cars pass each other at x = 44.5m, and if it has a constant velocity of 40 km/h, they pass each other at x=76.6m. What are (a) the initial velocity and (b) the constant acceleration of the green car?
How can the cars pass each other at 44.5 m and at 76.6m?
They start 220 m apart and move toward each other. The point where they pass depends upon the speed of the red car, for which two values are assumed.
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To find the initial velocity of the green car, we can use the fact that the cars pass each other at x = 44.5m and x = 76.6m for different speeds of the red car.
(a) Finding the initial velocity of the green car:
Let's first consider when the red car has a constant velocity of 20 km/h. At this speed, the red car travels a distance of 44.5m before passing the green car. We can calculate the time it takes for the red car to reach that point:
Distance = Velocity × Time
44.5m = (20km/h) × (Time in hours)
To find the time in seconds, we convert kilometers per hour to meters per second:
20 km/h = (20 × 1000m) / (60 × 60s) = 5.56 m/s
Now we can solve for the time:
44.5m = (5.56 m/s) × (Time in seconds)
Time = 44.5m / 5.56 m/s = 8 seconds
Since the green car is at x_g = 220m at t = 0, we can find its initial velocity using the equation of motion:
x_g = x_0 + v_0 * t + (1/2) * a * t^2
Substituting the known values:
220m = 0 + v_0 * 8s + (1/2) * a * (8s)^2
Simplifying the equation:
220m = 8s * v_0 + 32s^2 * a
Next, let's consider when the red car has a constant velocity of 40 km/h. At this speed, the red car travels a distance of 76.6m before passing the green car.
Using the same approach as before, we can find the time it takes for the red car to reach that point:
76.6m = (40km/h) × (Time in hours)
40 km/h = (40 × 1000m) / (60 × 60s) = 11.11 m/s
Now we can solve for the time:
76.6m = (11.11 m/s) × (Time in seconds)
Time = 76.6m / 11.11 m/s = 6.91 seconds
Using the equation of motion, we can find the initial velocity of the green car for this case as well:
220m = 0 + v_0 * 0s + (1/2) * a * (0s)^2
76.6m = 6.91s * v_0 + 23.77s^2 * a
(b) Finding the constant acceleration of the green car:
Now that we have two equations with two unknowns (v_0 and a), we can solve the system of equations to find their values.
We can write the equations as follows:
8s * v_0 + 32s^2 * a = 220m
6.91s * v_0 + 23.77s^2 * a = 76.6m
To solve the system of equations, we can use any method of simultaneous equations solving, such as substitution or elimination. In this case, we will use the elimination method.
Multiplying the first equation by 6.91 and the second equation by 8:
55.28s * v_0 + 221.12s^2 * a = 1517.2m
55.28s * v_0 + 190.16s^2 * a = 612.8m
Subtracting the second equation from the first equation, we eliminate v_0:
30.96s^2 * a = 904.4m
Now we can solve for a:
a = 904.4m / (30.96s^2) = 29.2m/s^2
Therefore, the initial velocity of the green car (a) is unknown, and the constant acceleration (b) of the green car is 29.2 m/s^2.