A car is moving at 40 mi/h along a straight line when the driver sees a light turn red. She hits the brakes when she is 100 ft from the light and decelerates at the rate of 16 ft per second squared. How far does she travel before stopping? Choose the whole number that is closest to the correct answer.
Hint: First convert 40 mi/hr into ft/sec.
Hint: See the Equations of Motion
do the first hint.
Vf^2=vi^2 + 2 ad solve for a.
25
To find how far the car travels before stopping, we need to use the equations of motion.
First, let's convert the car's speed from miles per hour (mi/h) to feet per second (ft/s).
1 mile = 5280 feet
1 hour = 3600 seconds
So, 40 mi/h = (40 * 5280) ft/ (1 * 3600) s = 176 ft/s
Now, let's use the equation of motion:
d = v0t + (1/2)at^2
Where:
d is the distance traveled
v0 is the initial velocity
a is the acceleration
t is the time
In this case, the initial velocity is 176 ft/s, the acceleration is -16 ft/s^2 (negative because it is decelerating), and we want to find the distance traveled (d).
Since the car is coming to a stop, its final velocity (vf) will be 0 ft/s.
So, plugging in the given values into the equation of motion:
0 = 176t + (1/2)(-16)t^2
Now, let's solve this quadratic equation for t.
0 = 176t - 8t^2
Rearranging, we have:
8t^2 - 176t = 0
Factoring out t:
t(8t - 176) = 0
Setting each factor equal to zero, we have two possible solutions:
t = 0 (which is the initial time when the car has not started decelerating)
or
8t - 176 = 0
Solving the second equation for t:
8t = 176
t = 176/8
t = 22
So, the car takes 22 seconds to come to a stop.
Now, let's calculate the distance traveled (d) using the equation of motion:
d = v0t + (1/2)at^2
d = 176 * 22 + (1/2) * (-16) * (22^2)
d = 3872 + (-176) * 242
d = 3872 - 42432
d = -38560
Since distance cannot be negative, we take the absolute value:
d = 38560 ft
Therefore, the car travels approximately 38560 feet before stopping.