To make a educational guess based on the your observations
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
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see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk
Net force on the car=F=4.8 x 10³ N
Explanation:
mass of car= 1.2 x 10³ Kg
initial velocity= Vi=0
Final velocity= Vf= 20 m/s
time = t= 5 s
Using kinematic equation,
Vf= Vi + at
20= 0 + a (5)
5 a=20
a= 20/5
a= 4 m/s²
Now force is given by F = ma
F= 1.2 x 10³ (4)
F=4.8 x 10³ N
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