distance traveled: .7*18+distance braking
distance braking: Vf^2=Vi^2+2ad or d=18^2/(2*4.6)=35.2m
distance trveled= .7*18+35.2m
a motorist traveling at 18m/s approaches traffic lights. when he is 30m from the stop line they turn red .it takes 0.7s before he can react by appling the breakes. the car slow down at a rate of 4.6m/s^2 how far from the stop line will he come to rest and on which side of it?
10 answers
nice but it is not cear
A distance travelled before applying a brake =0.7*18=12.6m
A distance travelled after applying brakes=vi2/2a,where as vf=0,s=sqrt18/2*4.6=35.2m
The total the motorist travel=12.6+35.2=47.8m
Therefore the distance between the stop line and the motorist stopped (the total motorist traveled) =47.8m-30m=17.8m
A distance travelled after applying brakes=vi2/2a,where as vf=0,s=sqrt18/2*4.6=35.2m
The total the motorist travel=12.6+35.2=47.8m
Therefore the distance between the stop line and the motorist stopped (the total motorist traveled) =47.8m-30m=17.8m
What about the side?
best
where is the side of the car and please draw the FBD
the side will be before the stope line or past the stop line
Vo=18m/s
Time to apply the brake is 0.7sec so, Distance travelled before he attempt to stop or apply the brake its S1=vot=18×0.7sec=12.6m
So when he applyed the brake its acceleration is 4.6 and so it stops where final velocity is 0m/s
Distance travelled after applying is caluculated by
Vsquare=initial square +2 acceleration ×distance two
So S2=324-0/2×4.6=35.2
The motorists goes 12.6m+35.2m =47.8m
That is a way greater than he should be that is 30 m its beyond that so it is 47.8-30=17.8m away from the stopline......which side of it.....he is out of the stopline that means away from the stop line...........example if he was coming from left side and stopline at the middle and traffic at the right......he is between the stopline and the traffic light that is to the right of the stopline
Time to apply the brake is 0.7sec so, Distance travelled before he attempt to stop or apply the brake its S1=vot=18×0.7sec=12.6m
So when he applyed the brake its acceleration is 4.6 and so it stops where final velocity is 0m/s
Distance travelled after applying is caluculated by
Vsquare=initial square +2 acceleration ×distance two
So S2=324-0/2×4.6=35.2
The motorists goes 12.6m+35.2m =47.8m
That is a way greater than he should be that is 30 m its beyond that so it is 47.8-30=17.8m away from the stopline......which side of it.....he is out of the stopline that means away from the stop line...........example if he was coming from left side and stopline at the middle and traffic at the right......he is between the stopline and the traffic light that is to the right of the stopline
Physics
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