A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 36 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)
No clue how to do it.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 36 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)
No clue how to do it.
To use the 2nd equation you need to know the inital and final velocity and the acceleration to find the time it takes to stop.
You are given explicity the initial velocity and the acceleration . But we want to solve for the time when the car is stopped so we know that
Using this you can solve for the time it takes for the car to stop.
After you get this information you can use all of the previous plus the new information to solve for the stopping distance. Also note for convience you can assume that .
I think its better to understand those formulas. Have you ever asked yourself where those formulas came from?
You do not need formulas for this problem, just a little bit logic.
First 1 foot = 0.000189(39) miles. So 1 mile is 1/0.000189(39)=5280 ft
Now 1 hour = 60 min.
1 min = 60 sec. so 60 min = 60*60sec=3600sec.
So the speed that you are traveling with is 50*5280ft/3600s = 73,(3) ft/s
Now you have a constant deceleration of 36 ft/s2.
What that really means? That means that the speed drops for 36 units every second.
Now how many seconds do we need so that the speed will drop to 0?
Now do some ratio.
1s____x s (seconds)
----=----
36____73 (drop of units ft/s)
(we can do this because of constant deacelleration)
Now just find x and tell me how many seconds does the car need so that it will stop?
You will tell me the feet that the car has covered.