A Driver takes 0.02 secs to apply brakes. If he is driving car at a speed of 54kmph and brakes causes deceleration of 6m/s2. Find the distance travelled by the car, after he sees the need to apply brakes..... Please tell how to solve it step by step too....
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Answered by
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the car is moving at speed of 54 km/hr =15 m/sec
i.e initial speed after applying brakes is (u) = 15 m/sec
distance covered is S₁ = v.t = 15*0.2 = 3 mtsfrom B to C it moves with constant deceleration of -6 m/sec2 and initial speed of 15m/secUsing relation V₂ - u₂ = 2*(-a)*SS₂ = V₂/2*a = 15*15 /(2*6) =18.75 mtsthe distance travelled by car after he sees need to put brakes is S1+S2= 21.75 mts =22 mts (approx)
i.e initial speed after applying brakes is (u) = 15 m/sec
distance covered is S₁ = v.t = 15*0.2 = 3 mtsfrom B to C it moves with constant deceleration of -6 m/sec2 and initial speed of 15m/secUsing relation V₂ - u₂ = 2*(-a)*SS₂ = V₂/2*a = 15*15 /(2*6) =18.75 mtsthe distance travelled by car after he sees need to put brakes is S1+S2= 21.75 mts =22 mts (approx)
ayushkumar17:
in s1 = 15 ×0.02 not 15 ×0.2 according to the qstn
Answered by
13
don't follow pic. follow this :-
speed =54km/h=15m/s
dist travel before applying brake = speed ×time
=15×0.02
=0.03m
to find dist travel after applying brake we use :-
v^2-u^2=2as
-u^2=2as(v=0)
15^2=-2×-6×s
s=225/12
=18.75m
total dist.= 18.75+0.03m
=19.05m
speed =54km/h=15m/s
dist travel before applying brake = speed ×time
=15×0.02
=0.03m
to find dist travel after applying brake we use :-
v^2-u^2=2as
-u^2=2as(v=0)
15^2=-2×-6×s
s=225/12
=18.75m
total dist.= 18.75+0.03m
=19.05m
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