Question

9.

A horse canters away from its trainer in a straight line,moving 160m away in 17s. It then turns abruptly and gallops halfwayback in 6.8s. Calculate the (a) its average speed and (b) itsaverage velocity for the entire trip, using "away from the trainer"as the positive direction.

(A) Average Speed

Formula:

(B) Average Velocity

Answer 1

Concepts and reason

The concepts used to solve this problem are average speed and average velocity.

Initially, the expression for the average speed is used to calculate the average speed for the entire trip.

Finally, the expression for the average velocity is used to calculate the average velocity for the entire trip.

Fundamentals

Average speed is defined as the rate of change of distance with time.

The expression for the average speed is given by,

$\bar s = \frac{{{d_{{\rm{tot}}}}}}{{{t_{{\rm{tot}}}}}}$

Here, $\bar s$ is the average speed, ${t_{{\rm{tot}}}}$ is the total time, and ${d_{{\rm{tot}}}}$ is the total distance.

Average velocity is defined as the rate of change of displacement with time.

The expression for the average velocity is given by,

$\bar v = \frac{{\Delta x}}{{\Delta t}}$

Here, $\Delta x$ represents the change in the displacement and $\Delta t$ represents the change in time.

(A)

The expression for the average speed $\bar s$ in terms total distance ${d_{{\rm{tot}}}}$ and the time ${t_{{\rm{tot}}}}$ is,

$\bar s = \frac{{{d_{{\rm{tot}}}}}}{{{t_{{\rm{tot}}}}}}$

The total distance is given by,

${d_{{\rm{tot}}}} = d + d'$

Here, d is the distance covered in 17 s and $d'$ is the distance covered in 6.8 s.

Substitute 160 m for d and 80 m for $d'$ in above equation as follows:

$\begin{array}{c}\\{d_{{\rm{tot}}}} = 160{\rm{ m}} + 80{\rm{ m}}\\\\ = 240{\rm{ m}}\\\end{array}$

The total time is given by,

${t_{{\rm{tot}}}} = t + t'$

Here, t is the time taken cover 160 m and $t'$is the time taken to cover half way distance.

Substitute 17 s for t and 6.8 s for $t'$ in above equation as follows:

$\begin{array}{c}\\{t_{{\rm{tot}}}} = 17{\rm{ s}} + 6.8{\rm{ s}}\\\\ = 23.8{\rm{ s}}\\\end{array}$

Substitute 240 m for ${d_{{\rm{tot}}}}$ and 23.8 s for ${t_{{\rm{tot}}}}$ in equation $\bar s = \frac{{{d_{{\rm{tot}}}}}}{{{t_{{\rm{tot}}}}}}$ as follows:

$\begin{array}{c}\\\bar s = \frac{{240{\rm{ m}}}}{{23.8{\rm{ s}}}}\\\\ = 10.08{\rm{ m/s}}\\\end{array}$

(B)

The expression for the average velocity $\bar v$is,

$\bar v = \frac{{\Delta x}}{{\Delta t}}$

The change in displacement is given by,

$\Delta x = {x_2} - {x_1}$

Here, ${x_1}$ and ${x_2}$are the initial and final positions.

Substitute 160 m for ${x_2}$ and 80 m for ${x_1}$, in above equation as follows:

$\begin{array}{c}\\\Delta x = 160{\rm{ m}} - 80{\rm{ m}}\\\\ = 80{\rm{ m}}\\\end{array}$

The change in time is given by,

$\Delta t = t + t'$

Here, t is the time taken cover 160 m and $t'$is the time taken to cover half way distance.

Substitute 17 s for t and 6.8 s for $t'$ in above equation as follows:

$\begin{array}{c}\\{t_{{\rm{tot}}}} = 17{\rm{ s}} + 6.8{\rm{ s}}\\\\ = 23.8{\rm{ s}}\\\end{array}$

Substitute 80 m for $\Delta x$ and 23.8 s for $\Delta t$ in equation $\bar v = \frac{{\Delta x}}{{\Delta t}}$ as follows:

$\begin{array}{c}\\\bar v = \frac{{80{\rm{ m}}}}{{23.8{\rm{ s}}}}\\\\ = 3.36{\rm{ m/s}}\\\end{array}$

Ans: Part A

The average speed for the entire trip is $10.08{\rm{ m/s}}$.