a car is moving with a speed of 10 metre per second when the braskes are applied the car has a constant negative acceleration (slows down) of -2m/s^2. what is its stopping distance?
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u = 10m/s
v = 0 m/s ( because the car applied brakes )
a = -2m /s ^2
Using third equation of motion
s = v^2 - u^2 / 2a
s = 0^2 - 10^2 / 2 ( -2 )
s = -100 / -4
s = 100/ 4
s = 25
Therefore distance travelled after applying brakes is 25m .
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v = 0 m/s ( because the car applied brakes )
a = -2m /s ^2
Using third equation of motion
s = v^2 - u^2 / 2a
s = 0^2 - 10^2 / 2 ( -2 )
s = -100 / -4
s = 100/ 4
s = 25
Therefore distance travelled after applying brakes is 25m .
Hope this helps ....Please mark me as Brainliest
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