1. ## tires

A car with tires that have a radius of 15 in. was driven on a trip and the odometer showed the distance traveled was 400 mi. With snow tires installed, the odometer showed that the distance for the return trip (same route) was 390 mi. Find the radius of the snow tires.

2. Originally Posted by Confused25
A car with tires that have a radius of 15 in. was driven on a trip and the odometer showed the distance traveled was 400 mi. With snow tires installed, the odometer showed that the distance for the return trip (same route) was 390 mi. Find the radius of the snow tires.
Let $\displaystyle n_1, r_1=15\mathrm{in}$ and $\displaystyle n_2, r_2$ be the number of turns of the tires and the radius of the tires in the two cases.
Since the distance travelled was the same in both cases, we have that
$\displaystyle \text{(I)}\quad n_1\cdot 2\pi r_1=n_2\cdot 2\pi r_2$
Also, the odometer deduces the distance travelled by counting the number of turns of the tires so that we have
$\displaystyle \text{(II)}\quad \frac{n_1}{n_2}=\frac{400\mathrm{mi}}{390\mathrm{m i}}$
From (I) we get that
$\displaystyle \text{(III)}\quad \frac{n_1}{n_2}=\frac{2\pi r_2}{2\pi r_1}=\frac{r_2}{r_1}$
Comparing (II) with (III) gives
$\displaystyle \frac{r_2}{r_1}=\frac{400\mathrm{mi}}{390\mathrm{m i}}$
Since $\displaystyle r_1$ is known, it is now easy to solve for $\displaystyle r_2$.

3. Originally Posted by Confused25
A car with tires that have a radius of 15 in. was driven on a trip and the odometer showed the distance traveled was 400 mi. With snow tires installed, the odometer showed that the distance for the return trip (same route) was 390 mi. Find the radius of the snow tires.
The distance travelled in one revolution is $\displaystyle 2\pi r$

The distance travelled in $\displaystyle k$ revolutions will $\displaystyle 2 \pi rk$

With your normal tyres this will be 400: $\displaystyle 2 \pi kr_1 = 400$ (eq1)

With snow tyres this will be 390: $\displaystyle 2 \pi kr_2 = 390$ (eq2)

$\displaystyle \frac{(eq2)}{(eq1)}$

$\displaystyle \frac{2 \pi k r_2}{2 \pi k r_1} = \frac{390}{400}$

$\displaystyle \frac{r_2}{15} = \frac{39}{40}$

Solve for $\displaystyle r_2$

4. I have two questions. Shouldn't snow tires have a bigger radius, not smaller?
Also, how can you assume that both tires travel the same revolutions?

5. Originally Posted by Confused25
I have two questions. Shouldn't snow tires have a bigger radius, not smaller?
Also, how can you assume that both tires travel the same revolutions?
Yes, you are right as regards the radius of the snow tires. $\displaystyle \mathrm{e}^{\mathrm{i}\pi}$ must have made a mistake. So try my answer (see above).
As regards the distance in both directions: we assume that the exact same route was travelled on the return trip.