The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life
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(a) To determine the least distance required to avoid a collision, we can use the equations of motion. The car is initially traveling at a velocity of 45 m/s and comes to rest under the influence of a friction force of 7.5 × 10^3 N.
We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (45 m/s)
a = acceleration (due to the friction force)
s = distance traveled
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0^2 - 45^2) / (2 * (-7.5 × 10^3))
s = 2025 / (-15 × 10^3)
s = -0.135 m
Since we can't have negative distance, the car needs to travel at least 0.135 m to avoid a collision with the other vehicle.
(b) If the initial distance between the vehicles is 40.0 m, we can find the speed at which the collision would occur. The car needs to come to rest within this distance.
Using the same equation as before:
s = (v^2 - u^2) / (2a)
We want to find the final velocity v.
0 = (v^2 - 45^2) / (2 * (-7.5 × 10^3))
Rearranging for v^2:
v^2 = 45^2
v = √(45^2) = 45 m/s
The collision would occur at a speed of 45 m/s.
(c) By solving these numerical problems, we can observe the importance of Physics in daily life. Physics helps us understand and analyze the motion of objects, the forces acting on them, and how these forces influence their behavior. In this case, Physics allows us to calculate the minimum distance required to avoid a collision and determine the speed at which a collision would occur. This knowledge can help drivers make informed decisions and take appropriate actions to ensure their safety and the safety of others on the road. Understanding the principles of Physics can also lead to the development of safer vehicles and improved road infrastructure. Overall, Physics plays a crucial role in various aspects of our daily lives, including transportation, technology, and safety.
To solve this problem, we need to use the equations of motion and the concept of force to calculate the answers. Let's go step by step.
First, let's address part (a) of the question: the least distance required to avoid a collision. To calculate this, we need to find the distance the car will travel during the deceleration period.
The equation we will use is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (0 m/s because the car comes to rest)
- vi is the initial velocity (45 m/s)
- a is the acceleration (in this case, it will be the deceleration caused by the friction force acting against the car's motion)
- d is the distance traveled
Rearranging the equation, we have:
d = (vf^2 - vi^2) / (2a)
Since the final velocity is 0 m/s, we can simplify it to:
d = -vi^2 / (2a)
Plugging in the values we have: vi = 45 m/s and a = 7.5 × 10^3 N, we can calculate the distance:
d = (-45^2) / (2 * 7.5 × 10^3)
= -2025 / (15000)
= -0.135 m
Now, since distance cannot be negative, we take the absolute value of the result:
d = 0.135 m
Therefore, the least distance the brakes should be applied to avoid a collision is 0.135 m.
Moving on to part (b) of the question: the speed at which the collision would occur. We need to find the final velocity of the red car when it covers the initial distance of 40.0 m.
To solve this, we'll use the equation of motion:
vf^2 = vi^2 + 2ad
Rearranging the equation and plugging in the values: vi = 45 m/s, a = 7.5 × 10^3 N, and d = 40.0 m:
vf^2 = (45^2) + 2 * (7.5 × 10^3) * 40.0
Simplifying the equation, we find:
vf^2 = 2025 + 600000
vf^2 = 602025
Taking the square root of both sides:
vf = √602025
= 775.74 m/s (rounded to two decimal places)
Therefore, the collision would occur at a speed of approximately 775.74 m/s.
Finally, for part (c) of the question, we can make some observations about the importance of physics in daily life based on this numerical:
1. Physics allows us to understand and predict the motion of objects. In this case, using equations of motion, we were able to determine the distance needed to avoid a collision and the speed at which a collision would occur.
2. Physics helps in designing and improving safety features in vehicles. Understanding concepts like force and motion allows engineers to develop efficient braking systems that can prevent accidents and protect lives.
3. Physics concepts, such as friction and Newton's laws of motion, are applied in various real-life scenarios, not just in this car collision example. From simple activities like walking to complex phenomena like space travel, physics plays a crucial role.
In conclusion, physics provides a foundation for understanding the fundamental principles of the natural world and enables us to apply this knowledge to solve practical problems in our daily lives.