a man driving at 20m/s get applied brake break which generated force to -500 November find distance after which the car is stop if it's mass is 100kg
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v = 20m/s
U = 0m/s
F = -500N
s= ?
m = 100kg
F = ma
-500 = 100a
a = -500/100 = -5m/s²
Using third equation of motion,
2as = v²-u²
2 X -5 X s = (20)² - 0²
-10s = 400 - 0
s = -40m
Car stops after travelling 40m
U = 0m/s
F = -500N
s= ?
m = 100kg
F = ma
-500 = 100a
a = -500/100 = -5m/s²
Using third equation of motion,
2as = v²-u²
2 X -5 X s = (20)² - 0²
-10s = 400 - 0
s = -40m
Car stops after travelling 40m
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