Constant speed and deaccelaration

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Homework Statement


R is traveling in a car at a constant speed of 18ms when she suddenly breaks.

if reaction time is .52s for applying foot to break how far is traveled before break is applied?

and if the break cause a deceleration of 5m/ssquared, how is traveled before stopping?
 
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ok sorry i not sure on the equations needed i have read the textbook. can you give me the equations so i can work them out? i thought mayb speed / time would give distance?
 
is it constant speed times time? if i do this tho my equation is 18 X .52 yes? 9.36 metres?
 
then the distance of deeceration is constant speed / 5mssquared?
 
sorry i don't mean 5mssquared i ment 5 ms per s
 
The constant speed of the car is 18 m/s and not 18 ms.

In case of constant speed the distance traveled is speed times time. Yes, the car traveled 9.36 meters before braking.
Constant acceleration means that the change of velocity is proportional to the time. (v2-v1)=at. In case of deceleration, v2<v1 and a is negative. You can get t from the change of velocity and a.
In case of accelerating motion, the displacement is
x= v0*t +a/2 *t^2.
Plug in the stopping time for t, and the appropriate value for a.

ehild
 
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?
 
access said:
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?

It is deceleration, so you need to plug in a=-5.

ehild
 
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
 
SteamKing said:
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
Not the car breaks, but the lady who drives it. :smile:

ehild