the time reaction for an automobile driver is 0.6 s if the automobile decelerates at 5m/s calcularte the total dist travelled in coming to stop from an initial velocity of 30km/h after a signal is observed
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Answered by
42
During the reaction time distance traveled = 0.6* 30*5/18 =5m
After observing and applying brakes:
s = (v^2 - u^2)/2a
=(0^2 - (30*5/18)^2) /(2*(-5)) m
s = 125/18 m
Total = 215/18 m
After observing and applying brakes:
s = (v^2 - u^2)/2a
=(0^2 - (30*5/18)^2) /(2*(-5)) m
s = 125/18 m
Total = 215/18 m
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Answered by
35
total distance traveled by particle = ut + u²/2a
where t is the time of reaction
u is the initial velocity
a is deacceleration of the body so,
distance = 30 *5/18*0.6 + (30*5/18)²/2*5
= 5 + 6.94
= 11. 94 m
where t is the time of reaction
u is the initial velocity
a is deacceleration of the body so,
distance = 30 *5/18*0.6 + (30*5/18)²/2*5
= 5 + 6.94
= 11. 94 m
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