Question

The figure shows the motion diagram for a horse galloping in one direction along a straight path. Not every dot is labeled, but the dots are at equally spaced instants of time. What is the horse's velocity during the first ten seconds of its gallop? What is the horse's velocity during the interval from 30s to 40s? What is the horse's velocity during the interval from 50s to 70s?

Answer 1

The expression for the velocity is defined as the displacement over time as follows:

\(v=\frac{d}{t}\)

From the given figure, For \(10 \mathrm{~s}:\) The change in position is the difference between final position to initial position is as follows:

$$ \begin{aligned} d &=500 \mathrm{~m}-600 \mathrm{~m} \\ &=-100 \mathrm{~m} \end{aligned} $$

Therefore, the average velocity at time \(10 \mathrm{~s}\) is,

$$ \begin{aligned} v &=\frac{-100 \mathrm{~m}}{10 \mathrm{~s}-0 \mathrm{~s}} \\ &=-10 \mathrm{~m} / \mathrm{s} \end{aligned} $$

For \(30 \mathrm{~s}\) to \(40 \mathrm{~s}:\)

The change in position is the difference between final position to initial position is as follows:

$$ \begin{aligned} d &=300 \mathrm{~m}-350 \mathrm{~m} \\ &=-50 \mathrm{~m} \end{aligned} $$

Therefore, the average velocity at time \(30 \mathrm{~s}\) to \(40 \mathrm{~s}\) is,

$$ \begin{aligned} v &=\frac{-50 \mathrm{~m}}{40 \mathrm{~s}-30 \mathrm{~s}} \\ &=-5 \mathrm{~m} / \mathrm{s} \end{aligned} $$

For \(50 \mathrm{~s}\) to \(70 \mathrm{~s}\) :

The change in position is the difference between final position to initial position is as follows:

\(d=50 \mathrm{~m}-250 \mathrm{~m}\)

$$ =-200 \mathrm{~m} $$

Therefore, the average velocity at time \(50 \mathrm{~s}\) to \(70 \mathrm{~s}\) is,

$$ \begin{aligned} v &=\frac{-200 \mathrm{~m}}{70 \mathrm{~s}-50 \mathrm{~s}} \\ &=-10 \mathrm{~m} / \mathrm{s} \end{aligned} $$