Calculating Deceleration: Swati's Story

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SUMMARY

Swati decelerated from 39 m/s to 27 m/s over a period of 6.0 seconds, resulting in a deceleration calculation using the formula a = Δv / Δt. The change in velocity (Δv) is 12 m/s, calculated as 39 m/s - 27 m/s. Substituting these values into the formula yields a deceleration of 2 m/s². This discussion clarifies the application of basic kinematic equations in real-world scenarios.

PREREQUISITES
  • Understanding of basic kinematics
  • Familiarity with the concept of acceleration and deceleration
  • Knowledge of the formula a = Δv / Δt
  • Ability to perform basic arithmetic operations with units
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  • Study the derivation and application of kinematic equations
  • Learn about the implications of acceleration in different contexts
  • Explore real-world applications of deceleration in automotive safety
  • Investigate the effects of friction and braking systems on deceleration
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Students studying physics, automotive engineers, and anyone interested in understanding the principles of motion and deceleration in vehicles.

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Homework Statement


Swati was driving her car over the speed limit when an oncoming car car flashes its lights warning her of a radar trap ahead. She decelerates from 39 m/s to 27 m/s in 6.0 s. What deceleration did her brakes apply?

- I am not sure what formula to use so does anyone have a clue what formula i could use for this question? Thanks for taking your time.:smile:
 
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Well, what's the definition of acceleration (or in this case deceleration) that you have come across in your lessons, or in your textbook?
 
umm well my teacher taught us this new displacement formula but i dunt think its related to the question i need help with .. but this is the formula he taught us: d= v1 deltat+1/2 a(deltat)2 sorry i dunt kno how to make the traingle for delta and the arrow heads... I am new to this
 
but we usually use the average velocity=delta d/delta t formula..
oh..i fthink its this formula a=delta t + v2-v1 is this correct?
 
Last edited:
\[<br /> a = {\textstyle{{\Delta v} \over {\Delta t}}}<br /> \]
 
thanks KingNothing
 
But if i was to use a=delta v/delta t .. how do i put to veclotiys? 39 m/s to 27 m/s
 
\Delta v means "change in velocity" just like \Delta t means change in time. Here, the change in time is 6 seconds, and the change in velocity is exactly what it sounds like it should be; namely (39-27)m/s.
 
oh so would i just do 39-27? which is 12 m/s?
 
  • #10
Yup, then substitute into the equation for acceleration.
 
  • #11
Alright, thanks a lot cristo for ur help and KingNothing for the formula. Thanks.
 

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