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Trajectory is the path of an object through space as a function of time. However, in many trajectory plots, when the movement is planar, a horizontal position axis and a vertical position axis are used instead. Yet, on these same plots, the instantaneous velocity vector is drawn tangent to the curve but velocity is the derivative of the position vector with respect to time so according to my logic, it is only tangent to the curve where the x-axis is time and the y-axis is position and not where the x-axis is horizontal position and the y-axis is vertical position.

Is plotting the instantaneous velocity vector tangent to a trajectory curve were axes are horizontal and vertical positions correct or is it just a negligence since the two plots are just a stretch of one another ? And if it is correct, why is that ?

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The answer is in the definition of instantaneous velocity. While moving along a path a small change in position ${\rm d}\vec{r}$ over a small time frame ${\rm d}t$ the instantaneous velocity is defined as $$\vec{v} = \frac{{\rm d}\vec{r}}{{\rm d}t}$$

The other way around, is that the change in position is

$${\rm d}\vec{r} = \vec{v}\, {\rm d}t$$

which means that since time is a scalar, that both ${\rm d}\vec{r}$ and $\vec{v}$ must point on the same direction.

It all comes down to the word along when describing the motion of particle. Since the motion is constrained/defined to be exactly on the path, it means that each small change ${\rm d}\vec{r}$ has to be tangent to the path.

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In a trajectory where $x = x(t)$ and $y = y(t)$, it is right that $v_x = x'(t)$ and $v_y = y'(t)$ are respectively tangent to the curves $x\times t$ and $y\times t$.

In a small neighborhood of a point $x_0 = x(t_0), y_0 = y(t_0)$, $$\frac{v_y}{v_x} = lim_{\delta t \to 0}\frac{\frac{y(t_0+\delta t) - y(t_0)}{\delta t}}{\frac{x(t_0+\delta t) - x(t_0)}{\delta t}} = lim_{\delta t \to 0} \frac{y(t_0+\delta t) - y_0}{x(t_0+\delta t) - x_0}$$ It is equivalent to the tangent to the trajectory: $$lim_{\delta x \to 0} \frac{y(x_0 + \delta x) - y(x_0)}{\delta x}$$

So, the velocity vector is tangent to the trajectory.

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