A car of mass 900 kg moves at 10 m/s. What is the braking force required to stop the car in a time of 5 seconds?
The average veloicity during braking is 5m/s, so in five seconds, it goes 25m.
brakingforce*5m=1/2 900 10^2
solve for braking force.
well if F=ma and a=v-u/t (final velocity-initial velocity/time) so combining these two equations we get F=m(v-u/t) which gives us F=mv-mu/t
if the car is to stop then obviously the final velocity is zero (now this will give us a negative answer but that is OK because it is a retarding force)
so our equation is F = (0-9000)/5
which = -1800
so therefor that is 1800 Newtons of force required to stop the car from moving.
If you have any questions ask away because i know this can get a little confusing.
To calculate the braking force required to stop the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the acceleration is the change in velocity divided by the change in time.
Given:
Mass of the car (m) = 900 kg
Initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s (since the car is stopping)
Time taken (t) = 5 seconds
First, we need to calculate the acceleration (a) using the equation:
a = (v - u) / t
Substituting the given values:
a = (0 - 10) m/s / 5 s
a = -10 m/s / 5 s
a = -2 m/s²
Since the car is decelerating (negative acceleration), the braking force will be in the opposite direction to the motion of the car.
Now, to calculate the braking force (F), we can use Newton's second law:
F = m * a
Substituting the values:
F = 900 kg * (-2 m/s²)
F = -1800 N
Therefore, the braking force required to stop the car in 5 seconds is 1800 Newtons in the opposite direction to its motion.