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Help on acceleration/ velocity!

 
 
pc6817
 
Reply Wed 14 Sep, 2005 11:35 pm
You are driving to the grocery store at 20 m/s. You are 110 m
from an intersection when the traffic light turns red. Assume
that your reaction time is 0.50 s and that your car brakes with
constant acceleration.
a. How far are you from the intersection when you begin to
apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take to stop?

Should i just use the following formulas? 1. Vf= Vi + at 2. Xf= Xi+ Vot+ (1/2) (a) (t^2) Thank you~
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vinsan
 
  1  
Reply Thu 15 Sep, 2005 03:12 am
Answer
When you are driving at 20 m/s in the reaction time of 0.5 seconds the distance covered will be 20*0.5 = 10 m

1. so you will apply breaks at 110-10 =100 m from traffic light

2. So now you have 20 m/s initial velocity and 100 m to cover

So the DECCELERATION should be to final velocity of 0 m/s

V= u + at

0 = 20 + at

at = -20 ... Formula 1


Also s = ut + (1/2)at^2

100 = 20t + (1/2)at^2 ... Formula 2

100 = 20t - 10t ... Substitute Formula 1

100 = 10t

t = 10


Put t = 10 in Formula 1

a = -2 m/s^2

It is negative as it is a DECCELERATION
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g day
 
  1  
Reply Thu 15 Sep, 2005 06:17 am
the simplest formulea when you have a distance an initial and final velocity and you want acceleration and don't care to know time is

V^2 - U^2 = 2aS so

0^2 - 20^2 = 2 a 100

0 - 400 = 200a

a = -2 m/s/s

And if you are slowing from 20 m/s sec at 2 m/sec/sec after an intial 0.1 lag - it will take you 0.5 + 20/2 = 10.5 seconds from the light change or 10 secs from when you hit the breaks - depending on what you wish to set as your frame of reference.
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