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A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s2 to his scooter. How far will it travel before stopping?

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प्रश्न

A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s2 to his scooter. How far will it travel before stopping?

 
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उत्तर

Initial velocity, u = 43.2 km/h = 12 m/s
Final velocity, v = 0
Acceleration, a = −6 m/s2
From \[v^2 = u^2 + 2\text{ as } \], we get:

Distance,

\[s = \frac{v^2 - u^2}{2(a)}\]
\[\Rightarrow s = \frac{0 - \left( 12 \right)^2}{2\left( - 6 \right)} = \frac{\left( 12 \right)^2}{12} = 12 \text{ m } \]
 
 

 

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पाठ 3: Rest and Motion: Kinematics - Exercise [पृष्ठ ५२]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 3 Rest and Motion: Kinematics
Exercise | Q 14 | पृष्ठ ५२

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