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

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