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

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The discussion focuses on calculating the necessary acceleration to stop a car right at an intersection when approaching at 20 m/s from 110 meters away. After a reaction time of 0.5 seconds, the car covers 10 meters, leaving 100 meters to the intersection. To find the required acceleration, the equation v^2 = u^2 + 2as is suggested, where the final velocity is 0 and the initial velocity is 20 m/s. The time to stop can be calculated using s = ut + 0.5at^2, utilizing the previously determined values. Participants confirm the calculations and equations, providing clarity on solving the problem effectively.
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 v^2=u^2+2as 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 s=ut+0.5at^2 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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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