A particle starts with velocity ( $u$ ) and moves with constant acceleration ( $a$ ). What is the nature of graph between the time ( $t$ ) and displacement ( $x$ )?

Straight line

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Symmetric parabola

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Asymmetric parabola

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Rectangular hyperbola

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Solution

The correct option is B Asymmetric parabola
$x = u t + \frac{1}{2} a t^{2}$ , is an equation for parabola.
It would have been symmetric, if $u = 0$ .
But, since $u \neq 0$ , it is asymmetric.

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Similar questions

u = 1 m s^{- 1}

a = 2 m s^{- 3}

x = u t + \frac{1}{2} a t^{2} = 1 \times t + \frac{1}{2} \times 2 \times t^{2} = t + t^{2}

(x) - t i m e (t)

t .

1

u

a

a

t = 0

t = t

u + x \times a t

4 x

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