The trajectory of a particle moving in a plan | vedclass

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(D) The average velocity vector between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{

Vedclass · Accepted Answer

$(4, 2)$

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(D) The average velocity vector between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}}{\Delta t}$.The slope of the secant line connecting these two points is $m = \frac{y_2 - y_1}{x_2 - x_1}$.For the points $A(0, 2)$ and $B(5, 3)$,the slope is $m = \frac{3 - 2}{5 - 0} = \frac{1}{5} = 0.2$.The instantaneous velocity at any point is given by the slope of the tangent to the trajectory at that point,$v_y / v_x = dy/dx$.We are looking for a point where the tangent slope is equal to the secant slope,i.e.,$dy/dx = 0.2$.Looking at the graph,at the point $(3, 1)$,the curve is transitioning from a downward slope to an upward slope,and the tangent is horizontal $(slope = 0)$.At the point $(4, 2)$,the tangent slope is positive and visually matches the slope of the line connecting $(0, 2)$ and $(5, 3)$.Thus,the correct coordinates are $(4, 2)$.

The trajectory of a particle moving in a plane is as shown in the figure. The coordinates of position $A$ are $(0, 2)$. The coordinates of another point at which the instantaneous velocity is the same as the average velocity between the points $(0, 2)$ and $(5, 3)$ are:

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