a car is running at 50 mph when the driver starts to step at the breaks. if the car decelerates at the rate of 7 ms^2, how long does it take the car stop?
Vo = 50mi/h * 1600m/mi * 1h/3600s. = 22.22 m/s.
V = Vo + a*t = 0 When stopped.
t = -Vo/a = -22.22/-7=3.17 s. To stop.
To calculate the time it takes for the car to stop, we can use the following equation:
v = u + at
where:
v is the final velocity (0 m/s, since the car stops),
u is the initial velocity (50 mph, which we'll convert to m/s),
a is the acceleration (negative 7 m/s^2, since the car is decelerating),
t is the time taken to stop.
First, let's convert the initial velocity from miles per hour (mph) to meters per second (m/s). We know that 1 mile is equal to 1609.34 meters and 1 hour is equal to 3600 seconds, so:
50 mph = (50 * 1609.34) / 3600 m/s
= 22.352 m/s (rounded to 3 decimal places)
Now we can plug the values into the equation:
0 = 22.352 m/s + (-7 m/s^2) * t
Rearranging the equation to solve for t:
-22.352 m/s = -7 m/s^2 * t
Dividing both sides by -7 m/s^2:
t = (-22.352 m/s) / (-7 m/s^2)
t = 3.193 seconds (rounded to 3 decimal places)
Therefore, it takes approximately 3.193 seconds for the car to come to a complete stop.