Calculating Initial Velocity and Acceleration of Passing Cars

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In summary: You have correctly converted the velocities from km/h to m/s. Your process for solving the problem is correct. However, there are a few mistakes in your calculations. For the first part, when the red car has a constant velocity of 20km/h, the time t is actually 8.01 seconds, not 2.225 seconds. And the final velocity of the green car is 21.91 m/s, not 44.5 m/s. For the second part, when the red car has a constant velocity of 40km/h, the time t is actually 6.894 seconds, not 1.915 seconds. And the final velocity of the green car is 20.80 m/s, not
  • #1
demonelite123
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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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  • #2
Before solving the problem, convert the velocity from km/h to m/s.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

1. How do you calculate the speed of the two passing cars?

The speed of the two passing cars can be calculated by dividing the distance between them by the time it takes for them to pass each other.

2. What factors affect the distance between the two passing cars?

The factors that affect the distance between the two passing cars include their initial speeds, the acceleration or deceleration of the cars, and any external forces such as wind or friction.

3. How does the mass of the cars affect the outcome of the passing cars problem?

The mass of the cars does not directly affect the outcome of the passing cars problem. However, it can indirectly affect the speed and acceleration of the cars, which in turn can affect the distance between them.

4. Can the passing cars problem be solved using Newton's laws of motion?

Yes, the passing cars problem can be solved using Newton's laws of motion, specifically the second law which states that the force applied to an object is equal to its mass multiplied by its acceleration.

5. How can the passing cars problem be applied in real-life situations?

The passing cars problem can be applied in real-life situations such as studying traffic flow and patterns, designing efficient transportation systems, and calculating the time and distance needed for safe passing on a two-lane road.

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