मराठी

A 1000 Kg Vehicle Moving with a Speed of 20 M/S is Brought to Rest in a Distance of 50 Metres : (I) Find the Acceleration. (Ii) Calculate the Unbalanced Force Acting on the Vehicle. - Science

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प्रश्न

A 1000 kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 metres :
(i) Find the acceleration.
(ii) Calculate the unbalanced force acting on the vehicle.

टीपा लिहा

उत्तर

(i) Final velocity, v = 0 (because the car is brought to rest)
Distance covered, s = 50 m
So, Force = mass × acceleration
But we do not know the value of a.
Making use of the third equation of motion we get,
v2 − u2 = 2as
⇒ 0 − 202 = 2 × a × 50
⇒ −400 = 100a
⇒ a = −4
The acceleration = −4 m/s2

(ii) Force = mass × acceleration
Mass = 1000 kg
Force  = 1000 × (−4)
           = − 4000 N
Unbalanced force acting on the vehicle is −4000 N.

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पाठ 2: Force and Laws of Motion - Very Short Answers 2 [पृष्ठ ७५]

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लखमीर सिंह Physics (Science) [English] Class 9 ICSE
पाठ 2 Force and Laws of Motion
Very Short Answers 2 | Q 36.2 | पृष्ठ ७५

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