Instantaneous Velocity

When we're moving in a car, the speedometer shows the speed at which the car is moving at that moment. The speed of an object at a single moment is referred to as "instantaneous velocity." As we have learned earlier, average velocity corresponds with average rate of change. Similarly, instantaneous velocity corresponds with instantaneous rate of change.

Thus, the instantaneous velocity at time \(t=a\) can be obtained by differentiating the position function \(x(t)\) with respect to time and plugging in \(t=a\), as shown below:

\[v(a)=\left.d\frac{x(t)}{dt}\right\rvert_{t=a}.\]

The position of a particle moving along the \(x\)-axis at time \(t\) can be described by

\[x(t)=3t^2-4t.\]

Find the particle's instantaneous velocity at \(t=2.\)

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Differentiating the position function and plugging in \(t=2\) gives

\[d\left.\frac{x(t)}{dt}\right\rvert_{t=2}=(6t-4)\big\rvert_{t=2}=12-4=8.\ _\square\]

The same goes for two-dimensional motion. Just consider the position and velocity as vectors, and apply the same rule.

The \(x\)- and \(y\)-coordinates of a particle moving on the \(xy\)-plane at time \(t\) are given by the following equations:

\[\begin{align} x(t)&=2t-5\\ y(t)&=-t^2+4t-1. \end{align}\]

Find the magnitude of the instantaneous velocity at \(t=1.\)

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Differentiating the position function with respect to \(t\) gives

\[\vec{v}=\left(d\frac{x(t)}{dt},d\frac{y(t)}{dt}\right)=(2,-2t+4).\]

Now, simply plugging in \(t=1\) gives the answer, as shown below:

\[\begin{align} \vec{v}(1) &=(2,-2\times1+4)\\ &=(2,2)\\\\ \Rightarrow \big\lvert\vec{v}(1)\big\rvert&=2\sqrt{2}.\ _\square \end{align}\]

Integration is the reverse process of differentiation. Hence, integrating the velocity function leads to the position function.

The positions of a particle moving along the \(x\)-axis with constant velocity at times \(t=1\) and \(t=3\) are \(x(1)=7\) and \(x(3)=15,\) respectively.

Find its position at \(t=14.\)

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Since the particle moves with constant velocity, its average velocity and instantaneous velocity are always equal. Thus its velocity function \(v(t)\) is a constant function. The particle's average velocity from \(t=1\) through \(t=3\) is

\[v(t)=\frac{x(3)-x(1)}{3-1}=\frac{15-7}{2}=4.\]

Integrating this with respect to \(t\) gives the position function, which is

\[x(t)=\int 4 dt=4t+C,\]

where \(C\) is the integration constant, and plugging in \(x(1)=1\) gives

\[x(1)=4\times1+C=7 \implies C=3 \implies x(t)=4t+3.\]

Therefore the position at \(t=14\) is

\[x(14)=4\times14+3=59.\ _\square\]