on turning a corner a car driver driving at 36 km per hour find the side on the road 55m ahead he immediately applies brakes so as to stop within 5 M of the child calculate the retardation produced and the time taken by the car to stop
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Solution
Initial velocity=u=36km/h= 36x(5/18)=10m/s
Final velocity=V=0 m/s acceleration=a=? time =t=? Distance travelled =S =55-5=50m By using Third equation of motion, V²-U²=2as ⇒0²-(10)²=2*a*50 a= -100/100 a= -1m/s². We know that, V=U+at 0=10-1xt -10=-1xt t=10sec =10 sec ∴ Acceleration is -1m/s² or retardation =1m/s² time take by the car to stop=10sec