# At what speed should it be traveling if the driver aims to arrive at Town B at 2.00 pm? [closed]

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?

-

## closed as off-topic by Bookend, Tunk-Fey, voldemort, user133281, Adam HughesSep 1 '14 at 6:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Bookend, Tunk-Fey, voldemort, user133281, Adam Hughes
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $480/7 \approx 68.6 \text{km/hr}$.

-

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.

-

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.

-