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Sagot :
Let the upstream breadcrumb's position act as the origin, so the position at time [tex]t[/tex] of upstream crumb [tex](B_U)[/tex], duck [tex](D)[/tex], and downstream crumb [tex](B_D)[/tex] are, respectively,
[tex]x_{B_U} = \left(2.0\dfrac{\rm m}{\rm s}\right) t[/tex]
[tex]x_D = d\,\mathrm m + v_D t[/tex]
[tex]x_{B_D} = 2d\,\mathrm m + \left(2.0\dfrac{\rm m}{\rm s}\right) t[/tex]
where [tex]d[/tex] is the distance (in m) between the duck and either crumb at the start, and [tex]v_D[/tex] is the velocity of the duck relative to the Earth.
To get both breadcrumbs, the duck has to choose between chasing the upstream or downstream crumb first.
• If the upstream crumb is chosen first, then [tex]v_D[/tex] is 2.0 - 1.0 = 1.0 m/s. Find the time it takes for the duck to reach the upstream crumb:
[tex]x_{B_U} = x_D \implies \left(2.0\dfrac{\rm m}{\rm s}\right) t = d + \left(1.0\dfrac{\rm m}{\rm s}\right) t \\\\ \implies t = d \, \mathrm s[/tex]
In this time, the duck covers a distance of
[tex]x_D = d\,\mathrm m + \left(1.0\dfrac{\rm m}{\rm s}\right) (d\, \mathrm s) = 2d \, \mathrm m[/tex]
while the downstream crumb will have traveled a distance of
[tex]x_{B_D} = 2d\,\mathrm m + \left(2.0\dfrac{\rm m}{\rm s}\right) (d \,\mathrm s) = 4d \,\mathrm m[/tex]
putting a total distance of [tex]2d+4d=6d\,\mathrm m[/tex] between the duck and the remaining crumb.
Take the duck's new position to be the new origin. When the duck turns around and travels with the current, its speed [tex]v_D[/tex] will be 2.0 + 1.0 = 3.0 m/s. Then at time [tex]t[/tex], the duck and remaining crumb have position relative to the new origin given by
[tex]x_D' = \left(3.0\dfrac{\rm m}{\rm s}\right) t[/tex]
[tex]x_{B_D}' = 6d \,\mathrm m + \left(2.0\dfrac{\rm m}{\rm s}\right) t[/tex]
Find the time the duck needs to catch up to the downstream crumb:
[tex]x_D' = x_{B_D}' \implies \left(3.0\dfrac{\rm m}{\rm s}\right) t = 6d\,\mathrm m + \left(2.0\dfrac{\rm m}{\rm s}\right) t \\\\ \implies t = 6d \,\mathrm s[/tex]
Then the total time the duck needs to get both crumbs is [tex]d + 6d = 7d \,\mathrm s[/tex].
• If instead the downstream crumb is chased down first, we start with [tex]v_D[/tex] = 3.0 m/s, so that
[tex]x_D = x_{B_D} \implies d\,\mathrm m + \left(3.0\dfrac{\rm m}{\rm s}\right) t = 2d\,\mathrm m + \left(2.0\dfrac{\rm m}{\rm s}\right) t \\\\ \implies t = d \, \mathrm s[/tex]
In this time, the duck covers a distance of
[tex]x_D = d\,\mathrm m + \left(3.0\dfrac{\rm m}{\rm s}\right) (d\,\mathrm s) = 4d \,\mathrm m[/tex]
while the upstream crumb moves
[tex]x_{B_U} = \left(2.0\dfrac{\rm m}{\rm s}\right) (d\,\mathrm s) = 2d \,\mathrm m[/tex]
The distance between the duck and remaining crumb is then [tex]4d-2d=2d\,\mathrm m[/tex]. Now the duck turns around with speed [tex]v_D[/tex] = 1.0 m/s. If the crumb's new position is taken to be the origin, then
[tex]x_{B_U}' = \left(2.0\dfrac{\rm m}{\rm s}\right) t[/tex]
[tex]x_D' = 2d\,\mathrm m + \left(1.0\dfrac{\rm m}{\rm s}\right) t[/tex]
Find the time it takes for the duck to get the last crumb:
[tex]x_{B_U}' = x_D' \implies \left(2.0\dfrac{\rm m}{\rm s}\right)t = 2d\,\mathrm m + \left(1.0\dfrac{\rm m}{\rm s}\right) t \\\\ \implies t = 2d\,\mathrm s[/tex]
Then the duck can get both crumbs in a total of [tex]d+2d=3d\,\mathrm s[/tex].
To get both crumbs in the shortest time, the duck should go after the downstream crumb first. This path over 2 times faster than the other.
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