Question

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

Answer 1

Answer: 3.06 m/ second square

Explanation: 126 km/hr = 126 * 1000/ 3600 = 35 m/s

as, v square = u square + 2as

0 = 35 * 35 + 2 * a * 200

   a = - 3.06 m/ second square

retardation = 3.06 m/ secong square

Answer 2

Answer:

Explanation:

Given: speed of the car: 126*5/18= 35m/s (since 1 hour= 3600 secs)

Displacement of the car: 200m

Formulae: v2 = u2 + 2as

sol =>  0 = (35*35) + 2*a*200

= 1225 + 400a = 0

= -1225/400 = a

= -3.0625 m/s2

time= v= u+at

-35= -3.0625m/s2 * t

-35/-3.0625 = t

11.4285714286 sec