Calculating Speed Needed to Pass Traffic Light at 15 m/s

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A car traveling at 15 m/s observes a traffic light 180 m ahead turning red, which stays red for 15 seconds. To pass the light as it turns green, the driver must adjust speed since maintaining 15 m/s would result in passing the light while it is still red. If the driver reduces speed to 12 m/s, they can reach the light just as it turns green. The discussion emphasizes the importance of understanding kinematic equations to solve for final velocity based on known variables. Ultimately, the problem highlights the need for a solid grasp of physics concepts rather than seeking direct answers.
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A car is traveling at 15 m/s when driver observes that a traffic light 180 m ahead of car turns red. The traffic light is timed for 15 seconds. If the car’s driver wishes to pass the light without stopping just as turns green again, what will be the speed of the car just it passes the light?


I have no clue for this question... can anyone give me a hint or tips? Thanks..
 
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Any1?... T.T
 
His speed will be the same. When the light turns red, it will take him 15 sec to reach the light. Because the light is timed for 15 seconds, he will not have to change speed at all.
 
If he doesn't change his speed, than he will be passing red light, not green light. because 15*15 = 225m so car would be already far past the red light on red light signal not green light signal..
 
HA my bad, hold on...
 
This is from grade 11 physics..
 
Can someone please help please
 
Now that I think about it, this is a bad question, because the driver could slow down or speed up until he reaches the light and just gun it right through there, so really, any speed would work.

I'm sure you're glad you heard that lol.
 
In any kinematic problem, if you know three of the five variables, you can find the fourth, or fifth using the correct equation.

You know, initial velocity, 15 m/s, displacement 180 m, and time, 15 s. You want to know final velocity, v. What formula features the relationship of those four variables.
 
  • #10
...if he reduces speed to 12 m/s right as the light turns red, he can maintain that speed until he reaches the light, and after 15 seconds, the light wouldve turned green again once he passes through.
 
  • #11
sweatband said:
...if he reduces speed to 12 m/s right as the light turns red, he can maintain that speed until he reaches the light, and after 15 seconds, the light wouldve turned green again once he passes through.

Thanks for the input, Sweat, but this problem is valid (although unrealistic), and there is a proper answer that is not 12 m/s. In kinematics, a uniform acceleration is assumed, and reality is put on the backburner.
 
  • #12
So then is it 180m= 15m/s*15s

But than i don't know how to solve this equation..
 
  • #13
Your other thread tells me not to answer you!
Anyways.. what all s-u-v-a-t formula you know? Then go through second last suggestion of "Chi Meson".
 
  • #14
rhxoehwhfh said:
So then is it 180m= 15m/s*15s

But than i don't know how to solve this equation..

As Chi mentioned... you know v1, you know t. you know d. now you need v2. which equation for uniform acceleration has the variables v1, v2, t and d?
 
  • #15
rhxoehwhfh said:
So then is it 180m= 15m/s*15s

But than i don't know how to solve this equation..

You are in serious trouble. Your initial post indicates you really don't know how to do anything. ("Hey, here are all the problems I have to do, someone do them for me.")

You subsequent post indicates you are mad that someone isn't doing all your work for you (on a Friday night).

This last post indicates that you can't even recognize the obvious next step to take once you are given perfectly clear, straightforward help. You don't need internet help, you need to crack open a textbook.

Goodbye.
 
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