Question
In a sudden stop, a car's seatbelt provides the force that stops the passenger. For a $90-\mathrm{kg}$ passenger, what's the seatbelt force if a car goes from highway speed of $100 \mathrm{~km} / \mathrm{h}$ to rest in $6.8 \mathrm{~s}$ ?
Step 1
We do this by multiplying the speed by 1000 (to convert kilometers to meters) and dividing by 3600 (to convert hours to seconds). This gives us: \[ u = 100 \times \frac{1000}{3600} = \frac{250}{9} \, \text{m/s} \] Show more…
Show all steps
Your feedback will help us improve your experience
Sanu Kumar and 64 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In an emergency stop to avoid an accident, a shoulder-strap seatbelt holds a 60 -kg passenger in place. If the car was initially traveling at $90 \mathrm{~km} / \mathrm{h}$ and came to a stop in $5.5 \mathrm{~s}$ along a straight, level road, what was the average force applied to the passenger by the seatbelt?
What average force is required to stop a $950-\mathrm{kg}$ car in $8.0 \mathrm{~s}$ if the car is traveling at $95 \mathrm{~km} / \mathrm{h} ?$
What average force is required to stop a 950-kg car in 8.0 s if the car is traveling at 95 km/h?
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD