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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, 10 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.

- 2 years, 10 months ago

Can be easily proved using $$v$$ $$Vs$$ $$t$$ graph

- 2 years, 10 months ago

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.

- 2 years, 10 months ago