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A car will travel from Town A to Town B. If it travels at a constant speed of 60 km/h, it will arrive at 3.00 pm. If travels at a constant speed of 80kh/h, it will arrive at 1.00 pm. At what speed should it be traveling if the driver aims to arrive at Town B at 2.00 pm?

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closed as off-topic by Live Forever, Tunk-Fey, voldemort, user133281, Adam Hughes Sep 1 '14 at 6:18

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Let $d$ be the distance between Town A and Town B. Let $x$ be a number so that 3.00PM - $x$ be the time that the driver started. We then have:

$$d = 60 \cdot x$$ $$d = 80 \cdot (x - 2)$$

Set the two equations equal to get:

$$ 80(x-2) = 60x $$ $$ 80x - 160 = 60x $$ $$ 20x = 160 $$ $$ x = 8 $$

Hence, the driver started at 8 hours before $3.00$. We want to find out how fast the driver should be if he wants to arrive after $x - 1$ hours ($3 - 2 = 1$). $x - 1 = 7$. Solving for $d$, we have $d = 480$.

$$ 480 = 7m $$

So the driver should drive at $ 480/7 \approx 68.6 \text{km/hr} $.

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The trip became $120$ minutes ($2$ hours) shorter by using $\frac34$ of a minute per kilometer ($80$ km/hr) instead of $1$ minute per kilometer ($60$ km/hr.) Since the savings from going faster was $\frac14$ of a minute per kilometer, the trip must be $480$ kilometers long, so it took $8$ hours at $60$ km/hr, and we set off at 7 AM. Therefore, to arrive at 2 PM, we should travel $480$ kilometers in $7$ hours, or $68\frac{4}{7}$ km/hr.

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Since both journeys are made at constant speeds the SUVAT equation $s = ut + \frac{1}{2}at^2$ (where $s$ measures displacement, $u$ is the initial velocity, $a$ the necessarily constant acceleration and $t$ the time) becomes $s=ut$. This is as we expect: if speed is constant then Distance = Speed $\times$ Time. Since the distances of the two journeys are equal we have $u_1t_1=u_2t_2$ where $u_i$ denotes the velocity of the $i^{\text{th}}$ journey in km/h and $t_j$ the time taken for the $j^{\text{th}}$ journey.

Let us assume that the first journey took $t_1$ hours. Since the second journey took two hours less, we have $t_2=t_1-2$. Thus: $u_1t_1 = u_2t_2$ becomes $60t_1 = 80(t_1-2)$ and hence $t_1=8$ hours. It also follows that $t_2 = 8-2 = 6$ hours.

For a third journey arriving at $2$ pm, we must have $t_3 = 7$ hours. Again, the distance is the same and so we have $u_1t_1=u_3t_3$ which becomes $60 \times 8 = 7u_3$, and hence $u_3=68\frac{4}{7}$ km/h.

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