a 1800 kg car is moving at 30m/S when breaks are applied if the average force exerted by the breaks is 6000 n find the distance travelled by the car before it comes to rest
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Answered by
33
F = ma
6000 = 1800 * a
a = 10 / 3 m/s^2
Using third equation of motion
v2 = u2 + 2as
0 = 30*30 + 2* 10/3 * s
-900 = 20s/3
s = -900 * 3 / 20
s = -135 m
Ignoring - sign
The car moved 135 m.
6000 = 1800 * a
a = 10 / 3 m/s^2
Using third equation of motion
v2 = u2 + 2as
0 = 30*30 + 2* 10/3 * s
-900 = 20s/3
s = -900 * 3 / 20
s = -135 m
Ignoring - sign
The car moved 135 m.
Answered by
10
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