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

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SUMMARY

The discussion revolves around solving a physics problem involving a speeder and a cop car, where the cop accelerates from rest at 9 km/h² until reaching a maximum speed of 170 km/h. The participants derive equations to find the time it takes for the cop car to catch the speeder, ultimately concluding that the total time required is approximately 26.79 seconds. The conversation also touches on related problems involving projectile motion and tangential acceleration, emphasizing the importance of understanding kinematic equations and conversions between units.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = v1*t + (1/2)at²
  • Ability to convert units between kilometers per hour and kilometers per second
  • Familiarity with solving quadratic equations
  • Knowledge of projectile motion concepts, including vertical and horizontal components
NEXT STEPS
  • Study the derivation and application of kinematic equations in various scenarios
  • Learn how to convert between different units of speed and acceleration
  • Practice solving quadratic equations and their applications in physics problems
  • Explore projectile motion equations, focusing on vertical and horizontal displacement
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and dynamics, as well as educators looking for examples of real-world applications of these concepts.

  • #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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