Physics, 02.06.2020 04:58 abelxoconda

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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Answer from: ladybugys

(a) The distance of the pig from the finish line is 428.3 m

(b) The lion was stationary for 4.32 s

The given parameters;

total distance covered by both animals, d = 2.20 km = 2,200 mspeed of the lion, v_l

= 18 m/sspeed of the pig, v_p

= 2.7 m/sinitial distance traveled by the lion, = 1.760 km = 1,760 m

When the pig meets the lion, both have covered a total distance of 1,760 m

The remaining distance to be covered = 2,200 m - 1,760 m = 440 m

Let the time both animals finished the remaining distance = t

Apply relative velocity concept; as the pig is moving, the lion is closing the gap between them until the finish line.

$(V_l - V_p)t = 440\\\\(18- 2.7)t = 440\\15.3 t = 440\\\\t = \frac{440}{15.3} \\\\t = 28.76 \ s$

If the lion had maintained a constant motion without stopping, the time it would have finished the remaining race is calculated as;

$t_l = \frac{440}{18} = 24.44 \ s$

(b) This show that the lion was stationary for (28.76 - 24.44 )s = 4.32 s

(a) The distance traveled by the pig during the 4.32 s that the lion was stationary is calculated as;

d = 4.32 x 2.7 = 11.7 m

Thus, the distance of the pig from the finish line is (440 - 11.7)m = 428.3 m.

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Answer from: batmandillon21

The lion covers a distance of $\left(18.0\frac{\rm m}{\rm s}\right)t$ after 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

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

Answer from: Quest

explanation:

could i give you a exaplo that could you?

Answer from: Quest

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

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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...

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