The displacement x of a particle in a straigh | vedclass

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(D) The displacement of the particle is given by the equation $x = 1 - t - t^2$.To find the velocity $v$,we differentiate the displacement with respec

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(D) The displacement of the particle is given by the equation $x = 1 - t - t^2$.To find the velocity $v$,we differentiate the displacement with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}(1 - t - t^2) = -1 - 2t$.To find the acceleration $a$,we differentiate the velocity with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(-1 - 2t) = -2 \, m/s^2$.Since the acceleration $a = -2 \, m/s^2$ is constant and negative,the $x-t$ graph must be a downward-opening parabola.At $t = 0$,$x = 1 - 0 - 0^2 = 1 \, m$. Thus,the graph starts at $x = 1$ on the vertical axis.At $t = 0$,the velocity $v = -1 - 2(0) = -1 \, m/s$. Since the velocity is negative,the slope of the graph at $t = 0$ must be negative.Among the given options,the graph that starts at $x = 1$ with a negative slope and curves downwards (representing constant negative acceleration) is Graph $D$.

The displacement $x$ of a particle in a straight line motion is given by $x = 1 - t - t^2$. The correct representation of the motion is $..............$.

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