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Acceleration doesn't matters?

A particle at rest starts moving with acceleration \(a_1\). Until it reaches the speed \(v\), it immediately decelerates with \(a_2\) until it stops. The total time consumed is \(t\). Show that the displacement of the particle is \(\frac{1}{2}vt\).

Note by Christopher Boo
2 years, 4 months ago

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It can also be interpreted as :
Draw the velocity vs time graph of the particle. We need to find the area under the graph which is the area of the triangle which simply is \( \dfrac{1}{2} \times \text{ base } \times \text{ height } = \dfrac{1}{2} \cdot t \cdot v \). This is the required value. Sudeep Salgia · 2 years, 4 months ago

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Can be easily proved using \(v\) \(Vs\) \(t\) graph Dinesh Chavan · 2 years, 4 months ago

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The average velocity with first is (0+v)/2=v/2, in second case it is (v+0)/2=v/2 same as first.
So d = (average velocity)*Time. Niranjan Khanderia · 2 years, 4 months ago

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