A car traveling at 72 mph starts braking and decelerates at a constant rate. the car takes 1/9 of a mile to stop. Find, to the nearest half-second, how long it took the car to cover the 1/9 of a mile.
I don't need the whole thing done. Just don't know how to start!
LOL
there is a hard way and an easy way
the easy way is to know that during constant acceleration the average speed is the average of the beginning and end speeds
average speed = 72/2 = 36 mph
speed * time = distance
36 * time = 1/9
time = 1 hour/(9*36)
1 hour = 3600 seconds
so in seconds
3600/(9*36) = 11 seconds
now the hard way is
v = Vi + a t
0 = 72 + a t
a = -72/t
x = Vi t + .5 a t^2
1/9 = 72 t + .5 (-72/t) (t^2)
1/9 = 72 t - .5(72 t)
or like we said before
1/9 = (72/2) t
To solve this problem, we can use the formula for distance traveled during deceleration:
d = (v^2 - u^2) / (2a)
Where:
- d is the distance traveled during deceleration
- v is the final velocity (0 mph in this case, as the car comes to a stop)
- u is the initial velocity (72 mph in this case, as that's when braking starts)
- a is the acceleration (you need to find this to solve the problem)
In this case, we know the initial velocity (u) is 72 mph, the final velocity (v) is 0 mph, and the distance traveled (d) is 1/9 of a mile. We need to solve for 'a' so we can then find the time taken to cover the distance.
Rearranging the formula, we get:
a = (v^2 - u^2) / (2d)
Plugging in the values we know:
- v = 0 mph
- u = 72 mph
- d = 1/9 mile
a = (0^2 - 72^2) / (2 * 1/9)
Simplifying further, we get:
a = (-72^2) / (2/9)
To find the acceleration (a), calculate (-72^2), then divide that value by 2/9.
Once you have the value of 'a', you can use it to calculate the time taken by using the formula:
t = (v - u) / a
Again, plugging in the values we know:
- v = 0 mph
- u = 72 mph
- a = the value you calculated in the previous step
This will give you the time taken by the car to cover the 1/9 mile distance. Round your answer to the nearest half-second for final result.