Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
62.5 m
Further explanation
Given:
- A 1000 kg car is traveling at 25 m/s (approx 55 mph).
- Its brakes can apply a force of 5000 N.
Question:
What is the minimum distance required for the car to stop?
The Process:
Newton's second law of motion: [tex]\boxed{ \ \Sigma F = ma \ }[/tex]
The car is said to experience braking force in the opposite direction to the car's movement. We will find out the amount of acceleration due to braking. In this issue, it is more precisely called deceleration.
[tex]\boxed{ \ \Sigma F = ma \ } \rightarrow \boxed{ \ a = \frac{- friction \ force}{m} \ }[/tex]
The minus sign on the frictional force indicates the opposite direction to the object motion.
[tex]\boxed{ \ a = \frac{- 5000}{1000} \ } \rightarrow \boxed{ \ a = - 5 \ m/s^2 \ }[/tex]
The minus sign emphasizes the acceleration in the form of deceleration. The car will stop after traveling a certain distance which we will find out.
Next, let us use one of the following equations of uniform motion.
[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex]
- v = final velocity → v = 0 m/s
- u = initial velocity → u = 25 m/s
- a = - 5 m/s²
- d = distance travelled
Let us calculate the minimum distance required for the car to stop.
0² = 25² + 2(-5)(d)
10d = 625
[tex]\boxed{d = \frac{625}{10}}[/tex]
Thus, the minimum distance required for the car to stop is 62,5 m.
- - - - - - - - - -
Alternative Steps
We can also use the relationship between work and the change in kinetic energy (ΔKE).
Recall that,
- work done (W) = force x distance,
- [tex]kinetic \ energy = \frac{1}{2}mv^2[/tex]
- and W = ΔKE.
Let us calculate the minimum distance required for the car to stop.
[tex]\boxed{ \ W = \Delta KE \ }[/tex]
[tex]\boxed{ \ -f \times d = \frac{1}{2}m(v^2 - u^2) \ }[/tex]
The minus sign on the frictional force indicates the opposite direction to the object motion.
[tex]\boxed{ \ -5000 \times d = \frac{1}{2}(1000)(0^2 - 25^2) \ }[/tex]
[tex]\boxed{ \ -5000 \times d = (500)(-625) \ }[/tex]
[tex]\boxed{ \ d = \frac{(500)(-625)}{-5000} \ }[/tex]
[tex]\boxed{ \ d = \frac{625}{10} \ }[/tex]
Thus, the minimum distance required for the car to stop is 62,5 m.
Learn more
- A case problem of uniformly accelerated motion and Newton's Second Law https://brainly.com/question/11181200
- Finding the acceleration between two vectors https://brainly.com/question/6268248
- Determine the flea's acceleration https://brainly.com/question/5424148
Keywords: friction forces, brakes, travelling, the minimum distance, Newton's Second Law, work, kinetic energy, velocity, uniformly accelerated motion, acceleration
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
