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Math Help - tires

  1. #1
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    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.
    Last edited by mr fantastic; August 7th 2009 at 11:38 PM. Reason: Restored deleted question.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Confused25 View Post
    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  n_1, r_1=15\mathrm{in} and 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
    \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
    \text{(II)}\quad \frac{n_1}{n_2}=\frac{400\mathrm{mi}}{390\mathrm{m  i}}
    From (I) we get that
    \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
    \frac{r_2}{r_1}=\frac{400\mathrm{mi}}{390\mathrm{m  i}}
    Since r_1 is known, it is now easy to solve for r_2.
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Confused25 View Post
    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 2\pi r

    The distance travelled in k revolutions will 2 \pi rk

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

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

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

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

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

    Solve for r_2
    Last edited by e^(i*pi); August 7th 2009 at 10:42 AM. Reason: got my constants messed up >.<
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  4. #4
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    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?
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Confused25 View Post
    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. \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.
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