How Long Does It Take for a Police Car to Catch a Speeder?

  • Thread starter Thread starter Sonic7z
  • Start date Start date
  • Tags Tags
    Kinematics
Click For Summary

Homework Help Overview

The problem involves a scenario where a speeder passes a parked police car, and the police car, starting from rest, accelerates to catch the speeder. The subject area pertains to kinematics, specifically the motion of objects under uniform acceleration and constant velocity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the different types of motion involved, noting the speeder's constant speed versus the police car's acceleration. There are suggestions to organize the information by creating separate headings for each vehicle and using appropriate equations for their respective motions.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the use of different equations for the two types of motion. There is an acknowledgment of the need to identify common factors in the motion of both vehicles to facilitate finding a solution.

Contextual Notes

Participants have pointed out potential issues with the equations initially presented, indicating that the original poster may need to clarify or correct their understanding of the motion involved.

Sonic7z
Messages
2
Reaction score
0

Homework Statement


A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44m/s2.

a. How much time passes before the speeder is taken over by the police car?
b. How far does the speeder get before bing overtaken?

Homework Equations


s=Vi+at
Vf=Vi+at
 
Physics news on Phys.org
Welcome to PF, sonic!
Careful with this one - there are two different kinds of motion involved. The speeder is moving at constant speed, so you need a simpler equation for the distance in that case. The equations you typed for accelerated motion are not quite right.

I suggest you make two headings, one for the speeder, one for the police car and write the appropriate equation(s) under each. Then fill in the numbers you know.

You'll need to identify what is the same for both cars and use that fact to put together a solution.

Good luck!
 
x=x0+v0t+1/2at^2

x=position
x0=initial value of position, here call it zero
v0=initial speed, 0 for the cop, 30m/s for the speeder
a=acceleration, 2.44m/s/s for the cop, 0 for the speeder


xcop=1/2*2.44m/s/s t^2

xspeeder = 30t

the cop catches the speeder when 1.22t^2 = 30t or when t=24.6s

in this time, the speeder travels 246.s x 30m/s = 738m
 
Looks great!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
7K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K