Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Peaceyou: An acceptance rate of 17% is terrible. You've asked eight questions and have left seven without feedback. If people take the time to write thoughtful, formatted replies then the least you could do is tick a box next to the most helpful answer. –  Fly by Night Jan 20 '13 at 2:02

3 Answers 3

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} $.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.