a car travelling with a speed of 72km/hr. the driver applied the breaks and retarded the car uniformly. the car is stopped in 5sec. find acceleration of the car, 2-distance before it stops after applying the break
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u = 72km/h × 5/18= 20m/sec
v= 0
time= 5 sec
acceleration = v-u/ t
= 0-20/5
= -20/5 = -4
so the acceleration of the car is 4m/sec^2 and the negative sign denotes it is retardation and the car is slowing down.
distance = ut + 1/2at^2
= 20×5 + 1/2 × -4 × 25
= 100 - 50
= 50 m
so distance travelled before it completely stops is 50m.
v= 0
time= 5 sec
acceleration = v-u/ t
= 0-20/5
= -20/5 = -4
so the acceleration of the car is 4m/sec^2 and the negative sign denotes it is retardation and the car is slowing down.
distance = ut + 1/2at^2
= 20×5 + 1/2 × -4 × 25
= 100 - 50
= 50 m
so distance travelled before it completely stops is 50m.
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