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?
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Scenes the chair wheels are up the person is rolling backwards and if the wheels were down then the person would go forwards
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Explanation:
Since its accelerating, the velocity vs time graph is linear
For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula
A= (vf - vi)/t
a= vf-0/t
1.25=vf / 7
1.25*7=vf
8.75 = vf
Now for displacement plug all the values in
X = 1/2(vf-vi)/t formula
The displacement (x) is 30.625 m
For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above
The answer is t = 5.02
Im pretty sure all the answers are correct