What acceleration will bring you to rest right at the intersection?

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SUMMARY

The discussion focuses on calculating the required acceleration to stop a car at an intersection when approaching at 20 m/s from a distance of 110 meters. After a reaction time of 0.5 seconds, the car travels 10 meters, leaving 100 meters to the intersection. The correct formula for acceleration is derived from the kinematic equation v² = u² + 2as, leading to the calculation of acceleration needed to stop the vehicle exactly at the intersection.

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  • Understanding of kinematic equations in physics
  • Familiarity with concepts of velocity and acceleration
  • Basic knowledge of motion under constant acceleration
  • Ability to perform algebraic manipulations to solve equations
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  • Study the kinematic equation v² = u² + 2as for motion analysis
  • Learn how to apply the equation s = ut + 0.5at² for distance calculations
  • Explore real-world applications of acceleration in automotive safety
  • Investigate the effects of reaction time on stopping distances in driving scenarios
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Students studying physics, automotive engineers, and anyone interested in understanding vehicle dynamics and stopping distances in traffic scenarios.

jelder
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You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50s and that your car brakes with constant acceleration.

How far are you from the intersection when you begin to apply the brakes?

What acceleration will bring you to rest right at the intersection?

How long does it take you to stop after the light changes?

the known values:
vi=20m/s
xi=110m
t=.50s
vf=0m/s
xf=0m
a=?

Now how would i solve using the kinematic equations. I've been spending quite a while on this one problem that seems easy but can't get the right answers?

1) I know there is equation for the first question but I did it in my head and got 100m. I new it was correct but didn't feel confortable because I know there had got to be an equation to find that answer


2) the equation I got to find acceleration: v-vi/t but I got 40m/s^2 and that is incorrect
 
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For the first part you know that you travel at 20m/s for 0.5 seconds, so you cover 10m using v=s/t, ie you're 100m from the lights when you apply the breaks, as you worked out.

For the second part, you can use [tex]v^2=u^2+2as[/tex] where v is the final velocity 0, u is the initial velocity 20, and s is the distance from the lights you worked out in part 1.

For the final part you can use [tex]s=ut+0.5at^2[/tex] where s and u are the same values from part 2, and the acceleration is what you work out from part 2.

Hope this helps!
 
thank you so much that helped a ton!
 

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