What Is the Initial Velocity of the Green Car with Constant Acceleration?

[cynical555]

Homework Statement

Two cars, a red one(left side) and a green one(right side), are moving toward each other in adjacent lanes. At time t=0, the red car is at position x= 0 m and the green car is at position x= 220 m. If the the red car has a constant velocity of 20km/hr, it passes the green car at x= 44.5 m. If the red car has a constant velocity of 40 km/hr, it passes the green car at x= 76.6 m. What is the initial velocity of the green car assuming its acceleration to be constant?

Homework Equations

The Attempt at a Solution

i found that the green car has moved 175.5 m after 8 s and that the car moved 143.4 m after 6.89s.
thanks in advanced!

 

[caljuice]

Now that you got the time, find the equation for the distance traveled for both cases and equate them together. Initial velocity and acceleration are the same for both of them, so you can find what you need. I got around ~51 m/s, ignoring sig figs. Hope that's correct.

 

[supratim1]

two unknowns - initial velocity (u) and acceleration (a) of green car. you get two equations of motions, and two unknowns. so solve them to get the answer.

 

[supratim1]

and Welcome to Physics Forums cynical555!

 

[rwisz]

Based on the information provided, we can apply the equations of motion to solve for the initial velocity of the green car assuming constant acceleration. The equations of motion are:

x = x0 + v0t + (1/2)at^2
v = v0 + at
v^2 = v0^2 + 2a(x-x0)

Where:
x = final position
x0 = initial position
v = final velocity
v0 = initial velocity
t = time
a = acceleration

Using the first set of data (red car velocity = 20 km/hr and green car position = 44.5 m), we can plug in the values and solve for the initial velocity of the green car:

44.5 m = 0 m + v0(8 s) + (1/2)a(8 s)^2
v0 = (44.5 m - 32s)/8 s
v0 = 3.06 m/s

Therefore, the initial velocity of the green car is 3.06 m/s.

Using the second set of data (red car velocity = 40 km/hr and green car position = 76.6 m), we can repeat the same process and solve for the initial velocity of the green car:

76.6 m = 0 m + v0(6.89 s) + (1/2)a(6.89 s)^2
v0 = (76.6 m - 23.6 s)/6.89 s
v0 = 7.26 m/s

Therefore, the initial velocity of the green car is 7.26 m/s.

To find the acceleration of the green car, we can use the second equation of motion and substitute in the values for the initial velocity found in the previous steps:

v = v0 + at
7.26 m/s = 3.06 m/s + a(6.89 s)
a = (7.26 m/s - 3.06 m/s)/6.89 s
a = 0.58 m/s^2

Therefore, the acceleration of the green car is 0.58 m/s^2. This means that the green car is accelerating at a constant rate of 0.58 m/s^2.

It is important to note that this solution assumes that the initial velocity of the green car is constant throughout the entire time period, which may not