A lion and a pig participate in a race over a 2.20 km long course. The lion travels at a speed of 18.0 m/s and the pig can do 2.70 m/s. The

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A lion and a pig participate in a race over a 2.20 km long course. The lion travels at a speed of 18.0 m/s and the pig can do 2.70 m/s. The lion runs for 1.760 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.
(a) How far (in m) is the pig from the finish line when the lion resumes the race? (b) For how long in time (in s) was the lion stationary?

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Ngọc Hoa 3 years 2021-09-04T19:47:30+00:00 1 Answers 0 views 0

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  1. kimchi2
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    2021-09-04T19:48:43+00:00
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    The lion covers a distance of \left(18.0\frac{\rm m}{\rm s}\right)t after t seconds, so it reaches the 1.760 km mark at time

    \left(18.0\dfrac{\rm m}{\rm s}\right)t=1760\,\mathrm m\implies t\approx97.8\,\mathrm s

    The pig travels a distance of \left(2.70\frac{\rm m}{\rm s}\right)t, so that it has moved

    \left(2.70\dfrac{\rm m}{\rm s}\right)(97.8\,\mathrm s)=264\,\mathrm m

    in the time it takes for the lion to move 1.760 km.

    (a) The lion has 0.44 km left in the race, which would take it

    \left(18.0\dfrac{\rm m}{\rm s}\right)t=440\,\mathrm m\implies t\approx24.4\,\mathrm s

    to finish.

    In order for the lion and pig to cross the finish line at the same time, the lion needs to resume running once the pig has 24.4 s remaining to the finish line; this happens when it is

    \left(2.70\dfrac{\rm m}{\rm s}\right)(24.4\,\mathrm s)\approx\boxed{65.9\,\mathrm m}

    away from the end.

    (b) The lion is stationary for as long as it takes the pig to cover the distance between [65.9 m away from the finish line] and [264 m from the starting line], or (2.20 km – 65.9 m) – 264 m = 1.87 km, which takes it

    \left(2.70\dfrac{\rm m}{\rm s}\right)t=1870\,\mathrm m\implies t=\boxed{935\,\mathrm s}

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