How far will a motorcycle travel before coming to a halt?

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SUMMARY

The discussion focuses on calculating the distance a uniformly retarded motorcycle travels before coming to a halt. The motorcycle covers 250 meters in the first 10 seconds and another 250 meters in the subsequent 20 seconds, leading to the determination of a retardation of 2.5 m/s². The equations used were d = ut + (1/2)at², applied separately for each time interval. A key point raised is the need to recognize that the initial velocity (u) differs between the two time periods, necessitating a relationship between them for accurate calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = ut + (1/2)at²
  • Knowledge of uniform acceleration and retardation concepts
  • Ability to manipulate algebraic equations to solve for variables
  • Familiarity with the relationship between distance, speed, and time
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to relate different initial velocities in uniformly accelerated motion
  • Explore graphical representations of motion to visualize speed vs. time
  • Investigate real-world applications of retardation in vehicle dynamics
USEFUL FOR

This discussion is beneficial for physics students, educators, and automotive engineers interested in understanding motion dynamics and the principles of uniform retardation in vehicles.

Kartik.
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1. A uniformly retarded moving motorcycle covers 250 meters in the first 10 seconds and and 250 meters in the next 20 seconds.How much will it travel more to come before coming to a halt?


250 = 10u - 50a , 250 = 20u - 200a. On solving these i get and a retardation of 2.5.
 
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Kartik. said:
1. A uniformly retarded moving motorcycle covers 250 meters in the first 10 seconds and and 250 meters in the next 20 seconds.How much will it travel more to come before coming to a halt?


250 = 10u - 50a , 250 = 20u - 200a. On solving these i get and a retardation of 2.5.



Although it is hard to understand your work, which was not well explained, I get that you applied the equation d = ut + (1/2)at2 twice, once for each time period. This is the correct thing to do.

Did you assume that 'u' was the same between both equations? Because it's not the same u. The u in your first equation is the initial velocity at the beginning of the 10 second period. The u in your second equation is the initial velocity at the beginning of the 20 second period. These are not the same, but you can *relate* them using the equation for speed vs time.
 
An alternative method to the one above, that gives you two equations with the same u (initial velocity) would be to use the fact that you know how far the bike has traveled after 30 seconds :)
 

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