Day out

Answer: 35 km/h

Using a speed-time graph

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$\therefore 5t = x-5$ and $xt=210$

$5t = x-5\Rightarrow t = \frac{x-5}5$

$xt = 210 \Rightarrow x\left(\frac{x-5}5\right)=210\

\hspace{18mm}\Rightarrow x(x-5)=1050\

\hspace{18mm}\Rightarrow x^2-5x-1050=0$

The quadratic formula gives: $$\begin{array}{rl}x& =\frac{5\pm \sqrt{25--4200}}{2}\\ & =\frac{5\pm \sqrt{169\times 25}}{2}\\ & =\frac{5\pm 65}{2}\end{array}$$ So $x=35$

Using time = distance $\div$ speed

Average speed over $210$ km is $x$ km/h, then time is $\dfrac{210}x$ hours.

Planned average speed is $x-5$ km/h, time at planned speed $\dfrac{210}{x-5}$ hours.

$1$ hour earlier than planned, so $\dfrac{210}x$ is $1$ less than $\dfrac{210}{x-5}$, so $$\begin{array}{rl}& \frac{210}{x}=\frac{210}{x-5}-1\\ \\ \Rightarrow & 210=\frac{210x}{x-5}-x\\ \\ \Rightarrow & 210(x-5)=210x-x(x-5)\\ \Rightarrow & 210x-1050=210x-x^2+5x\\ \Rightarrow & x^2-5x-1050=0\\ \Rightarrow & (x+30)(x-35)=0\end{array}$$ So $x=-30$ or $x=35$. Since $x>0, x=35$.

Using distance = speed $\times$ time

If Chris completed the journey in $t$ hours, distance $=x\times t$.

Chris intended to take $t+1$ hours.

He intended to travel $5$ km/h slower, so at an average speed of $x-5$ km/h.

Distance $=(x-5)(t+1)=xt-5t+x-5$

Distances are the same so $xt = xt-5t+x-5\Rightarrow0=-5t+x-5\Rightarrow x=5t+5$

And $xt=210$ so $(5t+5)t=210\Rightarrow (t+1)t=42$ so $t, t+1$ could be $6,7$ (or $-6, -7$)

$x=5\times6+5=35$