Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 3: EXERCISE 1A

The speed of sound in wood is $3300 \mathrm{m}{ \mathrm{s}}^{-1}$ and the speed of sound in air is $330 \mathrm{m}{ \mathrm{s}}^{-1}.$ A hammer hits one end of a $33 \mathrm{m}$ long plank of wood. Find the difference in time between the sound waves being detected at the other end of the plank and the sound being heard through the air.

An exercise routine involves a mixture of jogging at $4 \mathrm{m}{ \mathrm{s}}^{-1}$ and sprinting at $7 \mathrm{m}{ \mathrm{s}}^{-1}.$ An athlete covers $1 \mathrm{km}$ in $3$ minutes and $10$ seconds. Find how long she spent sprinting.

Two cars are racing over the same distance. They start at the same time, but one finishes $8 \mathrm{s}$ before the other. The faster one averaged $45 \mathrm{m}{ \mathrm{s}}^{-1}$ and the slower one averaged $44 \mathrm{m}{ \mathrm{s}}^{-1}.$ Find the length of the race.

Two air hockey pucks are $2 \mathrm{m}$ apart. One is struck and moves directly towards the other at $1.3 \mathrm{m}{ \mathrm{s}}^{-1}.$ The other is struck $0.2 \mathrm{s}$ later and moves directly towards the first at $1.7 \mathrm{m}{ \mathrm{s}}^{-1}.$ Find how far the first puck has moved when the collision occurs and how long it has been moving for.

A motion from point $A$ to point $C$ is split into two parts. The motion from $A$ to $B$ has displacement $s_1$ and takes time $t_1.$ The motion from $B$ to $C$ has displacement $s_2$ and takes time $t_2.$ 

Prove that if $t_1=t_2,$ the average speed from $A$ to $C$ is the same as the average of the speeds from $A$ to $B$ and from $B$ to $C.$

A motion from point $A$ to point $C$ is split into two parts. The motion from $A$ to $B$ has displacement $s_1$ and takes time $t_1.$ The motion from $B$ to $C$ has displacement $s_2$ and takes time $t_2.$ 

Prove that if $s_1=s_2,$ the average speed from $A$ to $C$ is the same as the average of the speeds from $A$ to $B$ and from $B$ to $C$ if, and only if, $t_1=t_2.$

The distance from point $A$ to point $B$ is $s.$ In the motion from $A$ to $B$ and back, the speed for the first part of the motion is $v_1$ and the speed for the return part of the motion is $v_2.$ The average speed for the entire motion is $v.$

Prove that $v=\frac{2v_1{v}_2}{v_1+v_2}.$

The distance from point $A$ to point $B$ is $s.$ In the motion from $A$ to $B$ and back, the speed for the first part of the motion is $v_1$ and the speed for the return part of the motion is $v_2.$ The average speed for the entire motion is $v.$

Deduce that it is impossible to average twice the speed of the first part of the motion; that is, it is impossible to have $v=2v_1.$