DESPERATE NEED OF HELP ON THIS Physics PROBLEM

  • Thread starter Thread starter everlasting_truth
  • Start date Start date
  • Tags Tags
    Desperate Physics
AI Thread Summary
The discussion revolves around a physics problem involving a police car and a speeder, where the police car accelerates after a delay. The key steps to solve the problem include converting speeds from km/h to m/s, calculating the initial distance the speeder gains during the police car's delay, and setting up equations for the distances traveled by both vehicles. Participants emphasize the importance of guiding the original poster (OP) to understand the problem rather than providing direct answers, adhering to forum guidelines. Ultimately, the solution involves using the quadratic formula to find the time it takes for the police car to overtake the speeder. The community aims to foster learning through hints and suggestions.
everlasting_truth
Messages
1
Reaction score
0
DESPERATE NEED OF HELP ON THIS Physics PROBLEM!

Okay for those who are really smart in Physics, please help! I've been staring at this problem for hours and I just can't figure it out. :rolleyes: I know that you need to find the final velocity and then time, but it just seems like there's not enough information. Well here's the problem:

An unmarked police car, traveling a constant 90 km/h, is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time elapses after the police car is passed until it over takes the speeder (assumed moving at constant speed)?

I don't need the answer or anything, I just need to know where to start off 'cause I'm totally stuck. If you figure it out please email, message, or even IM me on MSN at pesstimisticgurl23@yahoo.com.Thank you so much!
 
Last edited:
Physics news on Phys.org
Write an equation for the distance between the two cars. Start with the one second delay -- how far does that put the speeder ahead of the cop at the instant that the cop car begins its acceleration? And in that second phase (cop car accelerating), what is the equation for the relative distance between the two cars?
 
First, change km/h to m/s for convenience.
Then set up the equation x = Vit + (1/2)at^2 for each object:

Police car:
x + 13.889 = 25t + t^2 ------> x = 25t + t^2 - 13.889

Speeder:
x = 38.889t


Now, make x of police car and x of speeder equal each other and you get:
25t + t^2 - 13.889 = 38.889t ------> t^2 - 13.889t - 13.889


Solve for time by using the quadratic formula. This is the instance that the two objects meet each other.

Once you do the calculation, you get 14.8 second as an answer.
 
Last edited:
ysk1, We are not supposed to give out the answers to problems here in the Homework Forums. That is part of the guidelines that you agreed to in order to get into these HW forums. Our job is to give suggestions, hints and other tips to help the original poster (OP) to figure out the problem on their own. There is very little value to the OP to just post the answer. They need to figure out most of it on their own in order to learn. Please do not post the answer in any other threads in the Homework Forums.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top