What is the average braking force required of a 1200kg car in order for it to come to rest from 60km/h in a distance of 25m
Vi = 60 km/h (1 h/3600 s)(1000 m/km)
v = Vi + a t
0 = Vi + a t
t = -Vi/a
d = Vi t + (1/2) a t^2
25 = Vi (-Vi/a) + (1/2)a (-Vi/a)^2
25 = -Vi^2/a + (1/2)Vi^2/a
25 a = -(1/2)Vi^2
the kinetic energy it possesses
To calculate the average braking force required to bring a car to rest, we can use the following formula:
F = (m * v^2) / (2 * d)
Where:
F = braking force
m = mass of the car
v = initial velocity of the car
d = braking distance
Given:
m = 1200 kg (mass of the car)
v = 60 km/h = 16.7 m/s (convert to meters per second)
d = 25 m (braking distance)
First, we need to convert the kilometers per hour (km/h) to meters per second (m/s). We can do this by multiplying the value by 1000/3600:
v = 60 km/h * (1000 m / 1 km) * (1 h / 3600 s) ā 16.7 m/s
Now, we can calculate the average braking force using the formula mentioned above:
F = (m * v^2) / (2 * d)
= (1200 kg * (16.7 m/s)^2) / (2 * 25 m)
ā 1346 N
Therefore, the average braking force required for the 1200 kg car to come to rest from 60 km/h within a distance of 25 m is approximately 1346 Newtons.