A car moving along a straight highway with speed of 126 km h¯¹ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?
Given: The speed of the car is 126 kmh -1 and its stopping distance is 200 m .
The retardation of the car can be calculated using third equation of motion.
v 2 − u 2 = 2 a s
Where, u is the initial velocity of the car, v is the final velocity of the car, s is the stopping distance and a is the acceleration.
By substituting the given values in the above expression, we get
0 − ( 126 × 1000 3600 ) 2 = 2 × a × 200 a = − ( 126 × 1000 3600 ) 2 2 × 200 = − 3.06 ms − 2
Thus, the retardation of the car is 3.06 ms -2 .
The time taken by car can be calculated using first equation of motion.
v = u + a t
By substituting the given values in the above expression, we get
0 = 126 × 1000 3600 + ( − 3.06 ) t t = 126 × 1000 3600 3.06 = 11.4 s
Thus, the car takes 11.4 s to stop.
Given: The speed of the car is
The retardation of the car can be calculated using third equation of motion.
Where,
By substituting the given values in the above expression, we get
Thus, the retardation of the car is
The time taken by car can be calculated using first equation of motion.
By substituting the given values in the above expression, we get
Thus, the car takes
