To find the time taken by the car to stop, we can use the equation of motion:
v^2 = u^2 + 2as
where: v = final velocity (which is 0 m/s since the car comes to a stop) u = initial velocity a = acceleration (which is the negative value of the retardation) s = displacement
Converting the given speed from kilometers per hour (km/h) to meters per second (m/s):
54 km/h = 54,000 m/3600 s = 15 m/s (approximately)
Now, we can substitute the values into the equation of motion:
0^2 = (15)^2 + 2(-5)s
Simplifying the equation:
0 = 225 - 10s
10s = 225
s = 22.5 meters
Now, we have the displacement of the car (s = 22.5 m) and the acceleration (a = -5 m/s^2). We can use these values to find the time taken to stop by rearranging the equation of motion:
v = u + at
0 = 15 + (-5)t
5t = 15
t = 15/5
t = 3 seconds
Therefore, the car will take 3 seconds to stop when the brakes are applied with a retardation of 5 meters per second squared.