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I have a problem I haven't been able to solve for a class. A man took a trip in a car. He drove $70$ miles at a slower speed. Then, he went the next $300$ miles at a speed that was $40$ mph faster than earlier. The time he spent driving at the faster speed was twice the time spent at the slower speed. Find the two speeds. I think I could break this down to the following.

Let $s_1$ be the slower speed, $s_2$ the faster speed, $d_1$ the shorter duration, and $d_2$ the longer duration. Then

$d_1=\frac{70}{s_1}$

$d_2=\frac{300}{s_1+40}$

$d_2=2d_1$

and so

$2d_1=\frac{300}{s_1+40}$

I think in my head I can come up with $35$ mph and $75$ mph but that's just because I tried a bunch of numbers that seemed normal for driving. How can I solve this?

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  • $\begingroup$ When making substitutions, you should be looking for opportunities to eliminate variables. You got rid $d_2$, now use the first and fourth equations to eliminate $d_1$, leaving a single equation in $s_1$ $\endgroup$ Commented Nov 27, 2012 at 23:21

3 Answers 3

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Let $t_1$ be the time spent during the first part and $v_1$ be the speed of the first part

We have :

$t_1 \times v_1 = 70 $

$2t_1 \times (v_1 +40) = 300$

So by replacing $t_1 \times v_1$ in the second equation we deduce $ t_1 = 2 $.

And then we find $v_1 = 35$.

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Let $d_1$ be the time spent driving at a slower speed, and $s_1$ be the speed during that leg.

We can use your calculations: $$d_1=\frac{70}{s_1}\tag{1}$$ $$2d_1=\frac{300}{s_1+40}\tag{2}$$

Multiplying clearing the denominator in each of your equations gives:

$$d_1 \cdot s_1 = 70 \tag{1}$$

$$2d_1 \cdot (s_1 +40) = 2(d_1 \cdot s_1) + 40\cdot2d_1 = 300 \tag{2}$$

Substituting $(1)$ into $(2)$ gives us:

$$2(70) + 80d_1 = 140 + 80d_1=300$$

$$80d_1 = 300 - 140 = 160\quad \iff \quad d_1 = \frac{160}{80} = 2.$$

Using the value $d_1$ in equation $(1)$ to solve for $s_1$:

$$2s_1 = 70 \quad \iff \quad s_1 = \frac{70}{2} = 35\text{ mph}.$$

So your results are correct: $s_2 = s_1 + 40 = 35 + 40 = 75$ mph.

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time = distance/speed

Let s be the slower speed.

Solve the following equation for s:

2*(70/s) = (300/(s + 40))

s = 35 mph, s + 40 = 75 mph

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