Calculating Initial Velocity and Acceleration of Passing Cars

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The discussion revolves around calculating the initial velocity and acceleration of a green car moving toward a red car with known velocities. The red car's velocities are given as 20 km/h and 40 km/h, leading to different passing points, which are used to derive time and distance equations. The initial velocity of the green car is estimated at 20.80 m/s based on calculations, but there is confusion regarding the acceleration, which is found to be approximately 0.9946 m/s². The participants emphasize the need to consider the green car's acceleration in the equations to accurately solve for its initial velocity and acceleration. The method of solving the problem requires careful application of kinematic equations, particularly when both cars are in motion.
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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 x=0 and the green car is at x=220. 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 40km/h, they pass each other at x=76.6m. What are
a) the initial velocity of the green car?
and
b) the acceleration of the green car?

so i drew out a picture and said that 20t = 44.5, so t = 2.225s. also i said that 175.5=xt, so x = 78.88m/s.

then i used the next set of information and said 40t = 76.6, and found that t = 1.915 and therefore 143.4=xt so x = 74.88m/s.

am i supposed to get 2 different velocities? since 74.88 is smaller than 78.88, i said that 74.88 is the initial velocity.

for the acceleration, i did 78.88-74.44 / 2.225-1.915 = 12.90m/s^2

is my work correct? my textbook doesn't have the answer for this problem.
 
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Before solving the problem, convert the velocity from km/h to m/s.
 
rl.bhat said:
Before solving the problem, convert the velocity from km/h to m/s.

ok so 20km/h is 50/9 m/s and 40km/h is 100/9 m/s. i used the same process as above. 50/9 t = 44.5 so t = 8.01 and x = 21.91. 100/9 t = 76.6 so t = 6.894 and x = 20.80. so do i just say that the initial velocity of the green car is 20.80 m/s? basically i just want to know if my method for solving this problem is correct or not.

and also for acceleration i now get 21.91 - 20.80 / 8.01 - 6.894 = .9946 m/s2
 
The green car is moving with acceleration. When they meet, they must have traveled for the same time t.
So the first equation for green car will be
220 - 44.5 = 8.01*vo + 0.5*a*(8.01)^2 ...(1)
Similarly write down the second equation for the green car, when the red car has 40 km/h.
Then solve then two equation to find vo and a.
 
rl.bhat said:
The green car is moving with acceleration. When they meet, they must have traveled for the same time t.
So the first equation for green car will be
220 - 44.5 = 8.01*vo + 0.5*a*(8.01)^2 ...(1)
Similarly write down the second equation for the green car, when the red car has 40 km/h.
Then solve then two equation to find vo and a.

thanks! if you also can, can you tell me why my method is incorrect? how do you know which equation to use and how do you know when you are supposed to solve for certain variables first and then go use the main equation after? i understand your solution to solving the problem but i feel like i would never of thought of that. i have worked on this problem for at least an hour.
 
demonelite123 said:
ok so 20km/h is 50/9 m/s and 40km/h is 100/9 m/s. i used the same process as above. 50/9 t = 44.5 so t = 8.01 and x = 21.91. 100/9 t = 76.6 so t = 6.894 and x = 20.80. so do i just say that the initial velocity of the green car is 20.80 m/s? basically i just want to know if my method for solving this problem is correct or not.

and also for acceleration i now get 21.91 - 20.80 / 8.01 - 6.894 = .9946 m/s2
How did you get x = 21.91 and 20.80?What are they?
Green car is accelerating. In your solution you have not considered it.
 
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