Respuesta :
The lion covers a distance of [tex]\left(18.0\frac{\rm m}{\rm s}\right)t[/tex] after [tex]t[/tex] seconds, so it reaches the 1.760 km mark at time
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=1760\,\mathrm m\implies t\approx97.8\,\mathrm s[/tex]
The pig travels a distance of [tex]\left(2.70\frac{\rm m}{\rm s}\right)t[/tex], so that it has moved
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(97.8\,\mathrm s)=264\,\mathrm m[/tex]
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
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=440\,\mathrm m\implies t\approx24.4\,\mathrm s[/tex]
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
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(24.4\,\mathrm s)\approx\boxed{65.9\,\mathrm m}[/tex]
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
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)t=1870\,\mathrm m\implies t=\boxed{935\,\mathrm s}[/tex]
(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 m
- speed of the lion, [tex]v_l[/tex] = 18 m/s
- speed of the pig, [tex]v_p[/tex] = 2.7 m/s
- initial 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.
[tex](V_l - V_p)t = 440\\\\(18- 2.7)t = 440\\15.3 t = 440\\\\t = \frac{440}{15.3} \\\\t = 28.76 \ s[/tex]
If the lion had maintained a constant motion without stopping, the time it would have finished the remaining race is calculated as;
[tex]t_l = \frac{440}{18} = 24.44 \ s[/tex]
(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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