Answer: (a) \(t=5\). (b)(i) \(10\sqrt6-20\). (b)(ii) \(T=20\).

The displacement is

\(s=\dfrac{10t+100}{\sqrt{2t^2+100}}\).

Write this as

\(s=(10t+100)(2t^2+100)^{-1/2}\).

(a) Differentiate using the product rule:

\(\dfrac{\mathrm{d}s}{\mathrm{d}t}=10(2t^2+100)^{-1/2}+(10t+100)\left(-\frac12\right)(2t^2+100)^{-3/2}(4t)\).

So

\(\dfrac{\mathrm{d}s}{\mathrm{d}t}=10(2t^2+100)^{-1/2}-2t(10t+100)(2t^2+100)^{-3/2}\).

Put this over the common denominator \((2t^2+100)^{3/2}\):

\(\dfrac{\mathrm{d}s}{\mathrm{d}t}=\dfrac{10(2t^2+100)-2t(10t+100)}{(2t^2+100)^{3/2}}\).

The numerator is

\(20t^2+1000-20t^2-200t=1000-200t\).

Therefore

\(\dfrac{\mathrm{d}s}{\mathrm{d}t}=\dfrac{1000-200t}{(2t^2+100)^{3/2}}\).

At a maximum, \(\dfrac{\mathrm{d}s}{\mathrm{d}t}=0\), so

\(1000-200t=0\).

Hence

\(t=5\).

(b)(i) At \(t=0\),

\(s=\dfrac{100}{\sqrt{100}}=10\).

At the maximum, when \(t=5\),

\(s=\dfrac{10(5)+100}{\sqrt{2(5)^2+100}}=\dfrac{150}{\sqrt{150}}=\sqrt{150}=5\sqrt6\).

The particle starts at displacement \(10\), moves to the maximum displacement \(5\sqrt6\), and then returns to its starting point. So the total distance travelled is

\((5\sqrt6-10)+(5\sqrt6-10)=2(5\sqrt6-10)\).

Thus the total distance is

\(10\sqrt6-20\).

(b)(ii) The particle is back at its starting point when its displacement is again \(10\). Therefore set

\(\dfrac{10t+100}{\sqrt{2t^2+100}}=10\).

Divide by \(10\):

\(\dfrac{t+10}{\sqrt{2t^2+100}}=1\).

So

\(t+10=\sqrt{2t^2+100}\).

Squaring both sides gives

\((t+10)^2=2t^2+100\).

Expanding,

\(t^2+20t+100=2t^2+100\).

So

\(t^2-20t=0\).

Hence

\(t(t-20)=0\).

The solution \(t=0\) is the starting time, so the return time is

\(T=20\).

Answer: \(T=3\) and the acceleration during the last 4 seconds is \(-1.5\text{ m s}^{-2}\).

During the first \(T\) seconds, the acceleration is \(2\text{ m s}^{-2}\). Since the graph starts from rest, the velocity at time \(T\) is

\(2T.\)

The total distance travelled is the area under the velocity-time graph. From the diagram, this area is a trapezium with vertical height \(2T\), lower parallel side \(8\), and upper parallel side \(4-T\). Therefore

\(\dfrac12(8+4-T)(2T)=27.\)

Simplifying,

\(T(12-T)=27.\)

So

\(T^2-12T+27=0.\)

Factorising,

\((T-3)(T-9)=0.\)

From the diagram \(T\lt4\), so the valid value is

\(T=3.\)

In the last \(4\) seconds, the velocity decreases from \(2T\) to \(0\). Since \(T=3\), the velocity at the start of the last part is

\(2T=6\text{ m s}^{-1}.\)

The acceleration is the gradient of the velocity-time graph:

\(\dfrac{0-6}{8-4}=-\dfrac64=-1.5.\)

Therefore the acceleration during the last \(4\) seconds is

\(-1.5\text{ m s}^{-2}.\)

Answer: speed is the reflected velocity graph; acceleration is a straight line through the origin with positive gradient.

(a) Speed is the magnitude of velocity, so

\(\text{speed}=|v|.\)

This means that every part of the velocity-time graph above the \(t\)-axis stays unchanged. Any part below the \(t\)-axis is reflected in the \(t\)-axis, because negative velocity still gives positive speed.

So the speed-time graph should have the same shape as the velocity-time graph where \(v\geqslant0\), and the part where \(v\lt0\) should be drawn above the axis as its reflection.

(b) Acceleration is the gradient of the velocity-time graph:

\(a=\dfrac{\mathrm dv}{\mathrm dt}.\)

The velocity-time graph is part of a quadratic curve, so its gradient is a linear function of \(t\). Therefore the acceleration-time graph is a straight line.

The gradient of the velocity-time graph is zero when \(t=0\), so the acceleration-time graph passes through the origin. Since the quadratic curve opens upwards in the diagram, the gradient increases as \(t\) increases, so the acceleration-time graph is a straight line with positive gradient.