Physics 11 Speed and acceleraton

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SUMMARY

The discussion addresses a physics problem involving the deceleration of a car from a speed of 60.0 km/h, which equates to 16.667 m/s, with a constant deceleration of -8.0 m/s². The correct distance required to stop the car is 173.6 meters, achieved by using the average velocity during deceleration. The initial attempt incorrectly calculated the stopping distance by assuming constant speed, leading to an erroneous result of 347.24 meters. The solution emphasizes the importance of using the average velocity and the correct acceleration value in calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically v_f^2 = v_i^2 + 2ad
  • Knowledge of unit conversions, particularly between km/h and m/s
  • Familiarity with concepts of acceleration and deceleration in physics
  • Ability to calculate average velocity during uniform acceleration
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to convert between different units of speed and acceleration
  • Explore examples of average velocity calculations in deceleration scenarios
  • Investigate common pitfalls in physics problem-solving and how to avoid them
USEFUL FOR

Students studying physics, particularly those in high school physics courses, and anyone looking to improve their understanding of motion, speed, and acceleration calculations.

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[SOLVED] Physics 11 Speed and acceleraton

Homework Statement



The brakes on a car permit it to decelerate at the rate of -8.0m/s^2. How much distance is required to stop this car when it is traveling 60.0km/hr?

The answer given is 173.6m

Homework Equations



A=V/T
D=RT

The Attempt at a Solution


I solved it this way
60km/h = 16.667m/s
A=V/T 16.667m/s / -.80m/s^2 = 20.834 seconds
D=RT 16.667m/s x 20.834seconds = 347.24m

This is wrong and I do not understand why. Am I doing it right or is the answer wrong?
 
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The reason your answer is wrong is because you assumed the car maintained its 60.0 km/hr pace for the entire 20.834 seconds. It did not, because it was slowing down.

Since its deceleration is constant, you are justified in saying that the car traveled for 20.834 seconds at an average velocity of (16.667 + 0)/2 m/s, and you will come up with the right answer.

Another way to do this problem that avoids this potential pitfall is to just use the equation [itex]v_f^2 = v_i^2 + 2ad[/itex].
 
Adding to what Tedjn said, it was probably just a typo but make sure you use -0.8 ms^-2 for the acceleration or the answer is a tenth of what it's supposed to be. Your speed looks right.
 
Last edited:

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