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Standard VII

Physics

Displacement

A particle at...

Question

A particle at $x = 0$ and $t = 0$ starts moving along $+$ ve X-direction with velocity v then x varies with time as?

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t^{3}

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t^{2}

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t^{1 / 2}

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Solution

The correct option is A t
We know that, rate of change of displacement = velocity.
=> $\frac{d x}{d t}$ = v
dx= v.dt
Integrating both side, $\int_{o}^{x} d x = \int_{0}^{t} v . d t$
x-0= v(t-0)
x=vt
Hence, x varies with time linearly.

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A particle at x=0 and t=0 starts moving along +ve X-direction with velocity v then x varies with time as?

The correct option is A tWe know that, rate of change of displacement = velocity. => dxdt= vdx= v.dtIntegrating both side, ∫xodx=∫t0v.dtx-0= v(t-0)x=vtHence, x varies with time linearly.

A particle at $x = 0$ and $t = 0$ starts moving along $+$ ve X-direction with velocity v then x varies with time as?

The correct option is A t
We know that, rate of change of displacement = velocity.
=> $\frac{d x}{d t}$ = v
dx= v.dt
Integrating both side, $\int_{o}^{x} d x = \int_{0}^{t} v . d t$
x-0= v(t-0)
x=vt
Hence, x varies with time linearly.