a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 212 m. If the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 44.2 m. On the other hand, if the red car has a constant velocity of 46.0 km/h, they pass each other at x = 76.3 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.
44.2 + Dg = 212 m
Dg = 212 - 44.2 = 167.8 m = Distance of green car.
Vor = 23km/h = 23,000m/3600s = 6.39 m/s.
Vr = 46km/h = 46,000m/3600s = 12.78 m/s.
Vor*T1 = 44.2
6.39T1 = 44.2
T1 = 6.92 s. = Time to pass.
Vr*T2 = 76.3
12.78*T2 = 76.3
T2 = 5.97 s. = Time to pass.
a. Vog*T1 = 167.8 m.
Vog*6.92 = 167.8
Vog = 24.2 m/s.
76.3 + Dg = 212
Dg = 212-76.3 = 135.7 m.
Vg*T2 = 135.7
Vg*5.97 = 135.7
Vg = 22.7 m/s
b. a = (Vg-Vog)/(T2-T1) =
Solve for a.
To solve this problem, we can use the equations of motion relating to distance, velocity, time, and acceleration.
Let's start by calculating the time it takes for the cars to pass each other in both scenarios:
In the first scenario, where the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 44.2 m.
Using the equation: distance = velocity × time
44.2 m = 23.0 km/h × t
Converting km/h to m/s: 23.0 km/h = 23.0 × (1000/3600) m/s = 6.39 m/s
Rearranging the equation to solve for time:
t = 44.2 m / 6.39 m/s = 6.91 seconds
In the second scenario, where the red car has a constant velocity of 46.0 km/h, the cars pass each other at x = 76.3 m.
Using the same equation: distance = velocity × time
76.3 m = 46.0 km/h × t
Converting km/h to m/s: 46.0 km/h = 46.0 × (1000/3600) m/s = 12.78 m/s
Rearranging the equation to solve for time:
t = 76.3 m / 12.78 m/s = 5.97 seconds
Now that we have the time it takes for the cars to pass, we can calculate the initial velocity and constant acceleration of the green car.
(a) Initial velocity of the green car:
In both scenarios, the green car starts from xg = 212 m and reaches xr = 0 m.
Using the equation: velocity = displacement / time
For the first scenario:
Initial velocity = (0 m - 212 m) / 6.91 s = -30.70 m/s (negative sign indicates the green car is moving in the negative direction)
For the second scenario:
Initial velocity = (0 m - 212 m) / 5.97 s = -35.52 m/s
(b) Constant acceleration of the green car:
To find the acceleration, we need to use the kinematic equation: displacement = initial velocity × time + (1/2) × acceleration × time^2
For the first scenario: (using the given final position x = 44.2 m)
44.2 m = (-30.70 m/s) × 6.91 s + (1/2) × acceleration × (6.91 s)^2
Rearranging the equation and solving for acceleration:
acceleration = (44.2 m - (-30.70 m/s) × 6.91 s) / [(1/2) × (6.91 s)^2] = -2.78 m/s^2
For the second scenario: (using the given final position x = 76.3 m)
76.3 m = (-35.52 m/s) × 5.97 s + (1/2) × acceleration × (5.97 s)^2
Similarly, rearranging and solving for acceleration:
acceleration = (76.3 m - (-35.52 m/s) × 5.97 s) / [(1/2) × (5.97 s)^2] = -3.33 m/s^2
Therefore, the answers are:
(a) The initial velocity of the green car is -30.70 m/s in the first scenario and -35.52 m/s in the second scenario.
(b) The constant acceleration of the green car is -2.78 m/s^2 in the first scenario and -3.33 m/s^2 in the second scenario.
To solve this problem, we will use the equations of motion and the given information.
Let's first find the initial velocity of the green car.
Given:
- At t = 0, xr = 0 and xg = 212 m.
- The red car has a constant velocity of 23.0 km/h.
The equation for displacement is:
x = xi + vi*t + (1/2)*a*t^2
For the red car:
When x = 44.2 m:
44.2 = 0 + (23.0 km/h) * t + (1/2) * 0 * t^2 (as the red car has zero acceleration)
Converting the velocity to m/s:
v_red = 23.0 km/h * (1/3.6) m/s = 6.39 m/s
Substituting the values:
44.2 = 0 + 6.39 * t
t = 44.2 / 6.39 = 6.91 seconds
At t = 6.91 seconds, the green car is at xg = 212 m.
Therefore, the initial velocity of the green car is:
v_initial_green = (xg - xi) / t = (212 - 0) / 6.91 = 30.66 m/s
For the second scenario:
When x = 76.3 m:
76.3 = 0 + (46.0 km/h) * t + (1/2) * 0 * t^2
Converting the velocity to m/s:
v_red = 46.0 km/h * (1/3.6) m/s = 12.78 m/s
Substituting the values:
76.3 = 0 + 12.78 * t
t = 76.3 / 12.78 = 5.97 seconds
At t = 5.97 seconds, the green car is at xg = 212 m.
Again, the initial velocity of the green car is:
v_initial_green = (xg - xi) / t = (212 - 0) / 5.97 = 35.44 m/s
(a) The initial velocity of the green car is approximately +30.66 m/s for the first scenario and +35.44 m/s for the second scenario.
Now let's find the constant acceleration of the green car.
The equation for velocity is:
v = vi + a*t
For the green car:
v_green = v_initial_green + a*t
Substituting the values for the first scenario:
6.39 = 30.66 + a*6.91
a = (6.39 - 30.66) / 6.91 ≈ -3.69 m/s^2
Substituting the values for the second scenario:
12.78 = 35.44 + a*5.97
a = (12.78 - 35.44) / 5.97 ≈ -3.75 m/s^2
(b) The constant acceleration of the green car is approximately -3.69 m/s^2 for the first scenario and -3.75 m/s^2 for the second scenario.