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प्रश्न
The velocity-displacement graph of a particle is shown in figure.
- Write the relation between v and x.
- Obtain the relation between acceleration and displacement and plot it.
उत्तर
a. Consider the point P(x, v) at any time t on the graph such that angle ABO is θ such that tan θ = `(AQ)/(QP) = ((v_0 -v))/x = v_0/x_0`
When the velocity decreases from v0 to zero during the displacement, the acceleration becomes negative.
`v_0 - v = (v_0/x_0)x`
`v = v_0(1 - x/x_0)`
Is the relation between v and x.
b. `a = (dv)/(dt) = ((dv)/(dt))((dx)/(dx)) = ((dv)/(dx)) ((dx)/(dt))`
`a = (- v_0)/(x_0) v`
`a = ((v_0^2)/(x_0^2))x - (v_0^2/x_0)`
At `x = 0`
`a = (-v_0^2)/x_0`
At `a = 0`
`x = x_0`
The points are `(0, (-v_0^2)/x_0)` and `B(x_0, 0)`
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