A car travalling 90km/h, applies breaks with a force of 8400N to act on the 1050kg car.
a) what is the acceleration of the car when the breaks are applied?
b)how long does it take to stop.
c) how far does the car travel while it is breaking?
could anyone please help me with this, studying for exams and im not sure how to do this question.
F=ma or a=F/m
How long?
vfinal=vinit+ acceleration*time
notice acceleration is negative here.
How far?
distance=vinit*time + 1/2 a t^2
F=ma or a=F/m
How long?
vfinal=vinit+ acceleration*time
notice acceleration is negative here.
How far?
distance=vinit*time + 1/2 a t^2
To solve these questions, we'll use the equations of motion and Newton's second law of motion.
a) To find the acceleration of the car when the brakes are applied, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a).
Given:
Force (F) = 8400 N
Mass (m) = 1050 kg
Using the formula F = ma, we can rearrange it to solve for acceleration (a):
a = F / m
Substituting the given values, we have:
a = 8400 N / 1050 kg
a = 8 m/s^2
Therefore, the acceleration of the car when the brakes are applied is 8 m/s^2.
b) To find how long it takes for the car to stop, we need to use the equation of motion:
v_final = v_initial + (acceleration * time)
In this case, the car starts at an initial velocity (v_initial) of 90 km/h, which we need to convert to m/s:
v_initial = 90 km/h = (90 * 1000) / 3600 m/s = 25 m/s
Since the car eventually stops (v_final = 0 m/s), we can rewrite the equation as:
0 = 25 m/s + (8 m/s^2 * time)
Solving for time (t), we have:
25 m/s = 8 m/s^2 * time
time = 25 m/s / 8 m/s^2
time = 3.125 s
Therefore, it takes 3.125 seconds for the car to stop.
c) To find the distance the car travels while braking, we can use the equation of motion:
distance = v_initial * time + (1/2 * acceleration * time^2)
Using the known values:
v_initial = 25 m/s
acceleration = -8 m/s^2 (negative sign because the car is decelerating)
time = 3.125 s
Substituting these values into the equation, we have:
distance = 25 m/s * 3.125 s + (1/2 * -8 m/s^2 * (3.125 s)^2)
distance = 78.125 m - 39.0625 m
distance = 39.0625 m
Therefore, the car travels 39.0625 meters while braking.
I hope this helps you understand how to solve these types of problems. Remember to double-check your calculations and units to ensure accuracy. Good luck with your exams!