brake is applied to stop a car moving with a velocity of 36km/hr. If the retardation of the car is 5m/s². How long will it take to stop the car.
Answers
Answer:
Velocity of the car = 36 km/h = 10 m/s
Time in which car is to be stopped = 2 s
Final velocity of the stopped car = 0 m/s
Using the relation;
v = u + a t
and substituting v = 0 m/s; u = 10 m/s; and t = 2 s,
we get
0 = 10 + 2 a
=> a = - 10/2 = - 5 m/s².
Distance travelled by the car can be obtained using the concept of average velocity.
Averave velocity during the two second braking interval = ½ × (initial velocity at the beginning of the interval + final velocity at the end of the two second interval) = ⅓(10 m/s + 0 m/s) = 5 m/s
Distance travelled at this average speed in 2 second = 5 m/s ×2 = 10 m.
Answer:
Velocity of the car = 36 km/h = 10 m/s
Time in which car is to be stopped = 2 s
Final velocity of the stopped car = 0 m/s
Using the relation;
v = u + a t
and substituting v = 0 m/s; u = 10 m/s; and t = 2 s,
we get
0 = 10 + 2 a
=> a = - 10/2 = - 5 m/s².
Distance travelled by the car can be obtained using the concept of average velocity.
Averave velocity during the two second braking interval = ½ × (initial velocity at the beginning of the interval + final velocity at the end of the two second interval) = ⅓(10 m/s + 0 m/s) = 5 m/s
Distance travelled at this average speed in 2 second = 5 m/s ×2 = 10 m.
Explanation:
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