On turning a corner a car driver driving at 36km/h , finds a child on the road 55m ahead . He immediately applies brake , so as to stop within 5m of the child. Calculate the retardation produced and the time taken by the car to stop
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u=36 km/hr=36*5/18=2*5=10m/sec
v=0 m/sec
s=55m
v^2=u^2+2as
0^2=10^2+2*a*55
0=20+110a
-20=110a
a= -20/110= -2/11= -0.18m/sec^2
v=u+at
0=10+(-0.18)t
-10= -0.18t
10=0.18t
t=10/0.18=10*100/18=55.5 sec
So, a= -0.18m/sec^2
t=55.5sec
v=0 m/sec
s=55m
v^2=u^2+2as
0^2=10^2+2*a*55
0=20+110a
-20=110a
a= -20/110= -2/11= -0.18m/sec^2
v=u+at
0=10+(-0.18)t
-10= -0.18t
10=0.18t
t=10/0.18=10*100/18=55.5 sec
So, a= -0.18m/sec^2
t=55.5sec
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