numerical.-:a car is running at 54 km/h slows down to 36 km/h over a distance of 20 m.calculate:1.the retardation produced by its brakes.2.the time for which the brakes are applied . plzzzzz guys my holidays are finished i have 1 hr plzzzzzzz help me plzzzz
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u=54km/h = 54×5/18 = 15m/s
v=36km/h = 36×5/18 = 10m/s
s=20m
(1) v²=u²+2as
=> (10)²=(15)²+2×a×20
=> 100 = 225 + 40a
=> 100-225 = 40a
=> -125 = 40a
=> -125/40 = a
=> a = -3.125 m/s²
Retardation = 3.125 m/s²
(2) v=u+at
=> 10 = 15 = -3.125t
=> -5 = -3.125t
=> 5/3.125 = t
=> 1.6 sec = t
Time = 1.6 seconds
@:-)
u=54km/h = 54×5/18 = 15m/s
v=36km/h = 36×5/18 = 10m/s
s=20m
(1) v²=u²+2as
=> (10)²=(15)²+2×a×20
=> 100 = 225 + 40a
=> 100-225 = 40a
=> -125 = 40a
=> -125/40 = a
=> a = -3.125 m/s²
Retardation = 3.125 m/s²
(2) v=u+at
=> 10 = 15 = -3.125t
=> -5 = -3.125t
=> 5/3.125 = t
=> 1.6 sec = t
Time = 1.6 seconds
@:-)
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