A car moving along a straight highway with speed of 126 km/hr is broughtto astop within a distance of 200 m . Wht is the acceleration of the car (assumed uniform) and how long does it take for the car to stop.
Answers
Answer:
Time - distance upon time
200÷126=11.1//7
3.06 m/sec^2 and 11.4 Seconds
Given:
Initial velocity of car u = 126 km/hr
Converting into m/sec by multiplying with 5/18:
= 126 x 5/18
= 35 m/sec
Distance traveled s = 200 meters
Final velocity = 0
To Find:
Retardation of the car
Time taken by the car to stop
Calculating:
For solving this question we use the equation of motion for finding acceleration of the car:
v^2 - u^2 = 2as
Substituting the values which are known to us in this formula we get:
(0)^2 - (35)^2 = 2 a (200)
- 1225 = 400 a
Taking 400 to the other side we get:
a = - 1225 / 400
a = - 3.06 m/sec^2
Therefore, the acceleration of the car assumed uniform is - 3.06 m/sec^2
Now to calculate time taken we use the formula of motion:
v = u + at
Substituting the values known to us in this formula we get:
0 = 35 + (- 3.06) t
- 35 = - 3.06 x t
Taking - 3.06 to the other side we get:
t = - 35 / - 3.06
t = 11.4 seconds
Therefore, the time taken for the car to stop is 11.4 seconds.