# Mike's Car Stopping Force Problem: Find the Answer!

• Visual1Up
In summary, the conversation discusses finding the average force exerted by seatbelts on a 90kg passenger as a car is stopped from a velocity of 15m/s in 2.0m by a mound of dirt. The correct formula to use is Vf ^ 2 = Vi ^2 + 2a DeltaX, and the correct value for acceleration is -56.25m/s^2. This results in a force of approximately 5100N on the passenger, which is the correct answer.
Visual1Up
Hey, I have the following problem:

A car going 15m/s is brought to rest in a distance of 2.0m as it strikes a mound of dirt. How large an average force is exerted by the seatbelts on a 90kg passenger as the car is stopped.

I have found one formula... F = (mass * change in velocity)/time ... and I have found stopping time to be approx (2.0/15) = 1.33 (unless I am wrong) so F = (90 (not sure about this) * 15)/1.33 = 10125N

The only problem is I have the right answer from a friend that is approx 5100N. Can someone tell me what's going wrong? Thanks!,
-Mike

Visual1Up said:
Hey, I have the following problem:

A car going 15m/s is brought to rest in a distance of 2.0m as it strikes a mound of dirt. How large an average force is exerted by the seatbelts on a 90kg passenger as the car is stopped.

I have found one formula... F = (mass * change in velocity)/time ... and I have found stopping time to be approx (2.0/15) = 1.33 (unless I am wrong) it's T = d/v_avg; v_avg is not 15, what should it be?, solve for T and watch your decimal pointso F = (90 (not sure about this) * 15)/1.33 = 10125N

The only problem is I have the right answer from a friend that is approx 5100N. Can someone tell me what's going wrong? Thanks!,
-Mike

As he had said, 15 is not the Vavg. Instead of using v to calculate, you should be using acceleration. Use the formula Vf ^ 2 = Vi ^2 + 2a DeltaX . Vf will be zero(because it comes to rest), Vi is 15, and delta X(displacement) is 2. Through the calculation, we get -56.25m/s^2 for acceleration. Force is mass times acceleration. The passenger's mass is 90kg, so we multiply the two figures, and get 5062.5. Because there's 2 significant figures, we round the number to 5100N.

Last edited:

## 1. What is the "Mike's Car Stopping Force Problem?"

The "Mike's Car Stopping Force Problem" is a hypothetical physics problem that asks for the amount of force needed to stop a moving car in a specific distance.

## 2. How is the problem solved?

The problem is solved using Newton's second law of motion, which states that force equals mass times acceleration (F=ma). This formula is used to calculate the force needed to stop the car in the given distance.

## 3. What factors affect the stopping force?

The main factors that affect the stopping force include the mass of the car, the speed of the car, and the surface on which the car is stopping. Friction and air resistance may also play a role in the calculation.

## 4. Can the problem be solved in real life?

Yes, the problem can be solved in real life using the same principles of physics. However, the exact force needed to stop a car may vary depending on real-world variables such as the condition of the car's brakes and tires, road conditions, and the driver's reaction time.

## 5. Why is this problem important?

This problem is important because it helps us understand the relationship between force, mass, and acceleration, which are fundamental concepts in physics. It also has practical applications, such as in designing safe braking systems for vehicles and understanding the physics of car accidents.

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
658
• Introductory Physics Homework Help
Replies
3
Views
930
• Introductory Physics Homework Help
Replies
9
Views
3K
• Classical Physics
Replies
46
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K