a car is moving at a rate of 90 km per hour and applies brakes which provide a retardation of 6 m per second how much time does the car takes to stop how much distance does the car cover before coming to rest what would be the stopping distance needed if speed of the car is doubled
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Initial velocity of car ,
u = 90 km/h = 90 * 5/18=25 m/s
Final velocity v=0
S= 25 m
Using third equation of motion ,
v^2 - u^2 = 2 a s
0 - 625 = 2 a *25
-625 = 2 a * 25
a = -625/2 * 25= - 12.5 m / sec square
u = 90 km/h = 90 * 5/18=25 m/s
Final velocity v=0
S= 25 m
Using third equation of motion ,
v^2 - u^2 = 2 a s
0 - 625 = 2 a *25
-625 = 2 a * 25
a = -625/2 * 25= - 12.5 m / sec square
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