Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 6: EXERCISE 1C

A driver sees the traffic lights change to red $240 \mathrm{m}$ away when he is travelling at a velocity of $30 \mathrm{m}{ \mathrm{s}}^{-1}.$ To avoid wasting fuel, he does not brake, but lets the car slow down naturally. The traffic lights change to green after $12 \mathrm{s},$ at the same time as the driver arrives at the lights.

What assumptions have been made to answer the question?

In a game of curling, competitors slide stones over the ice at a target $38 \mathrm{m}$ away. A stone is released directly towards the target with velocity $4.8 \mathrm{m}{ \mathrm{s}}^{-1}$ and decelerates at a constant rate of $0.3 \mathrm{m}{ \mathrm{s}}^{-2}.$ Find how far from the target the stone comes to rest.

A golf ball is struck $10 \mathrm{m}$ from a hole and is rolling towards the hole. It has an initial velocity of $2.4 \mathrm{m}{ \mathrm{s}}^{-1}$ when struck and decelerates at a constant rate of $0.3 \mathrm{m}{ \mathrm{s}}^{-2}.$ Does the ball reach the hole?

A driverless car registers that the traffic lights change to amber $40 \mathrm{m}$ ahead. The amber light is a $2 \mathrm{s}$ warning before turning red. The car is travelling at $17 \mathrm{m}{ \mathrm{s}}^{-1}$ and can accelerate at $4 \mathrm{m}{ \mathrm{s}}^{-2}$ or brake safely at $8 \mathrm{m}{ \mathrm{s}}^{-2}.$ What options does the car have?

The first two equations are $v=u+at$ and $s=\frac{1}{2}\left(u+v\right)t$ . You can use these to derive the other equations. By substituting for $v$ in the second equation, derive $s=ut+\frac{1}{2}a{t}^2.$

The first two equations are $v=u+at$ and $s=\frac{1}{2}\left(u+v\right)t.$ You can use these to derive the other equations.Derive the remaining two equations, $s=vt-\frac{1}{2}a{t}^2$ and $v^2=u^2+2as,$ from the original two equations.

Show that an object accelerating with acceleration $a$ from initial velocity $u$ to final velocity $v,$ where $0<u<v,$ over a time $t$ is travelling at a velocity of $\frac{u+v}{2}$ at time $t=\frac{1}{2}t$, that is at the time halfway through the motion the velocity of the object is the mean of the initial and final velocities.

Show that an object accelerating with acceleration $a$ from velocity $u$ to velocity $v,$ where $0<u<v,$ over a displacement $s$ is travelling at a speed of $\sqrt{\frac{v^2+u^2}{2}}$ at a distance $\frac{1}{2}s.$ Hence, prove that when the object does not change direction the speed at the midpoint of the distance is always greater than the mean of the initial and final speeds. Deduce also that the mean of the initial and final speeds occurs at a point closer to the start of the motion than the end.