A car is going 18.3 m/s down a road. At time t=0s, a squirrel runs in front of the car 36 meters away. It takes the driver t=0.67s to slam on the brakes, but after that , the brakes offer a frictional force F(fr)=21000N. Her car masses at 1860 kg.
a) How far does the car travel, in meters, before the driver hits the brakes?
b)How much time passes between stepping on the brakes and the car coming to a stop?
Let's solve the problem step by step:
a) To find the distance the car travels before the driver hits the brakes, we need to calculate the time it takes for the driver to notice the squirrel and react. We can use the equation:
distance = velocity × time
Given that the initial velocity of the car is 18.3 m/s and the time it takes for the driver to react is 0.67 seconds, we can substitute these values into the equation:
distance = 18.3 m/s × 0.67 s
distance = 12.261 meters
Therefore, the car travels 12.261 meters before the driver hits the brakes.
b) To find the time it takes for the car to come to a stop after the driver applies the brakes, we need to calculate the deceleration of the car using the equation:
force = mass × acceleration
Since the force exerted by the brakes is given as 21000 N and the mass of the car is 1860 kg, we can substitute these values into the equation:
21000 N = 1860 kg × acceleration
Solving for acceleration:
acceleration = 21000 N / 1860 kg
acceleration ≈ 11.29 m/s^2
From this, we can use the kinematic equation to calculate the time it takes for the car to come to a stop:
final velocity^2 = initial velocity^2 + 2 × acceleration × distance
Since the final velocity is 0 m/s (the car comes to a stop), the initial velocity is 18.3 m/s, and the distance the car travels is unknown (let's call it x):
0^2 = 18.3^2 + 2 × 11.29 × x
0 = 334.89 + 22.58x
22.58x = -334.89
x ≈ -14.8 meters
Since we're dealing with distances, we can discard the negative value. Therefore, the positive solution indicates that the car travels approximately 14.8 meters after the driver hits the brakes.
To calculate the time it takes for the car to come to a stop, we can use the equation:
time = final velocity / acceleration
Since the final velocity is 0 m/s and the acceleration is 11.29 m/s^2:
time = 0 / 11.29
time ≈ 0 seconds
Therefore, the car comes to a stop immediately after the driver applies the brakes (approximately 0 seconds pass between stepping on the brakes and the car coming to a stop).
To answer these questions, we need to apply the equations of motion and the concepts of friction and retardation.
a) How far does the car travel, in meters, before the driver hits the brakes?
Before the driver hits the brakes, the car is moving at a constant velocity of 18.3 m/s. We need to find the distance traveled during this time.
The formula to calculate distance is: distance = speed × time
Given that the car is moving at a speed of 18.3 m/s, and the time required to hit the brakes is not mentioned, it is assumed that the driver hits the brakes at t = 0 seconds.
Therefore, the distance traveled before hitting the brakes is:
Distance = speed × time
= 18.3 m/s × 0 seconds
= 0 meters
So, the car does not travel any distance before the driver hits the brakes.
b) How much time passes between stepping on the brakes and the car coming to a stop?
To calculate the time it takes for the car to come to a stop after stepping on the brakes, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
Net force = mass × acceleration
In this case, the net force acting on the car is the frictional force provided by the brakes (F(fr) = 21000 N), and the acceleration is the deceleration or retardation experienced by the car.
The formula to calculate deceleration is:
Deceleration = Net force / mass
Deceleration = F(fr) / mass
= 21000 N / 1860 kg
Now, we have the acceleration.
To calculate the time required for the car to come to a stop, we can use the equation of motion:
final velocity = initial velocity + (acceleration × time)
As the final velocity is 0 m/s (since the car comes to a stop), and the initial velocity is 18.3 m/s (the velocity before braking), we can substitute these values into the equation:
0 = 18.3 m/s + (acceleration × time)
We need to solve for time, so let's rearrange the equation:
time = (0 - 18.3 m/s) / acceleration
Substituting the value of acceleration we calculated earlier, we get:
time = (0 - 18.3 m/s) / (21000 N / 1860 kg)
Simplifying further, we get:
time = -18.3 m/s × (1860 kg / 21000 N)
Finally, evaluating the expression, we find:
time ≈ -1.62 s
It's important to note that the negative sign indicates that the car stops in the opposite direction that it was initially moving. Therefore, we can ignore the negative sign, and the time it takes for the car to come to a stop after stepping on the brakes is approximately 1.62 seconds.