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ad-work [718]
3 years ago
10

A spaceship starting from a resting position accelerates at a constant rate of 9.8 m/s. How far will the spaceship travel if its

final speed is 1 percent of the speed of light (300, 000, 000 m/s?)
Physics
1 answer:
Dvinal [7]3 years ago
3 0

300 000 0 squared = 2 x 9.8 distance

KINEMATICS

Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.


speed = distance/time

velocity = displacement/time

s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time


v=u+at

v^2=u^2+2as

s=ut+1/2at^2

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1 year ago
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 19
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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
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=> -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:

brainly.com/question/8898885

#SPJ1

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