Solve Car Chase Question: 110 km/h and 170 km/h

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Homework Help Overview

The discussion revolves around a physics problem involving a speeder and a cop car, where the speeder travels at a constant speed of 110 km/h and the cop car accelerates from rest at 9 km/h² until it reaches a maximum speed of 170 km/h. Participants are trying to determine how long it takes for the cop car to catch up with the speeder, with various attempts to relate their distances and times.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equal distances traveled by both vehicles and explore the necessary formulas to relate their speeds and times. There are questions about how to handle the conversion of units and the implications of acceleration on distance. Some participants suggest calculating the time it takes for the cop car to reach its maximum speed and the distance covered during that time.

Discussion Status

There is ongoing exploration of the problem with various participants offering insights and calculations. Some have provided partial calculations and suggestions for further steps, while others express confusion about unit conversions and the overall approach. Multiple interpretations of the problem are being considered, and participants are actively engaging with each other's ideas.

Contextual Notes

Participants are grappling with the conversion of speeds from km/h to km/s and the implications of acceleration on the distances traveled. There is also a mention of the initial conditions of the problem, including the cop car starting from rest and the need to determine the initial distance between the two vehicles.

  • #61
anglum said:
so then for the horizontal distance it is just 17 cos 50 (2.62)?

yes.
 
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  • #62
ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?
 
  • #63
ok i got that problem on my own actually WOOO HOOO

now the 2nd problem if u don't mind
 
  • #64
014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.
015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.
 
Last edited:
  • #65
learning these are my last 2 problems ... sorry to keep bothering you
 
  • #66
anglum said:
014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.

This is just the change in velocity divided by time. ie 0.958/1.95.

015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

What is the centripetal acceleration?
 
  • #67
anglum said:
ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

use centripetal acceleration = v^2/r. At a radius r, v = rw.
 

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