A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200m.What is the retardation of the car and how long does it take for the car to stop?​

• Speed of car = 126 km/h × 5/18 =35 m/s
• Distance = 200 m

• Retardation
• Distance to stop

According to 3rd equation of motion

v² - u² = 2as

0² - 35² = 2(a)(200)

0 - 1225 = 400a

-1225 = 400a

a = -1225/400

a = -0.3625 m/s ²

Now,

According to 1st equation of motion

v = u + at

0 = 35 + -0.3625 t

0 - 35 = -0.3625t

-35 = -0.3625t

t = -35/-0.3625

11.4 = t

Hence, Time = 11.4 sec