The x-t graph for a uniformly accelerated par | vedclass

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Question

(B) For a uniformly accelerated particle,the instantaneous velocity at any point is given by the slope of the tangent to the $x-t$ graph at that point

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) For a uniformly accelerated particle,the instantaneous velocity at any point is given by the slope of the tangent to the $x-t$ graph at that point,i.e.,$v = 	an 	heta$.At point $(i)$,the angle is $45^{\circ}$,so the velocity $v_1 = 	an 45^{\circ} = 1 \ ms^{-1}$.At point $(ii)$,the angle is $	heta_2$ such that $	an 	heta_2 = 3$,so the velocity $v_2 = 3 \ ms^{-1}$.For motion with uniform acceleration,the average velocity between two points is the arithmetic mean of the instantaneous velocities at those points:$v_{	ext{average}} = \frac{v_1 + v_2}{2}$$v_{	ext{average}} = \frac{1 + 3}{2} = \frac{4}{2} = 2 \ ms^{-1}$.

The $x-t$ graph for a uniformly accelerated particle is as shown in the figure. Find the average velocity between points $(i)$ and $(ii)$ in $ms^{-1}$.

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