Respuesta :
Answer:
distance difference would a) increase
speed difference would f) stay the same
Explanation:
Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.
Their equations of motion for distance and velocities are
[tex]s_2 = gt^2/2[/tex]
[tex]s_1 = g(t + \Delta t)^2/2[/tex]
[tex]v_2 = gt[/tex]
[tex]v_1 = g(t + \Delta t)[/tex]
Their difference in distance are therefore:
[tex]\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2[/tex]
[tex]\Delta s = g/2((t + \Delta t)^2 - t^2)[/tex]
[tex]\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t)[/tex] (As[tex]A^2 - B^2 = (A-B)(A+B)[/tex]
[tex]\Delta s = g\Delta t/2(2t + \Delta t)[/tex]
So as time progress t increases, Δs would also increases, their distance becomes wider with time.
Similarly for their velocity difference
[tex]\Delta v = v_1 - v_2 = g(t + \Delta t) - gt[/tex]
[tex]\Delta v = gt + g\Delta t - gt = g\Delta t[/tex]
Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.
This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.
Answer:
The velocity of the first diver is always more than the velocity of the second diver hence more distance covered by the first diver over the same time
Explanation:
In free fall
v^2 = u^2 + 2×g×S
If the initial velocity of the first diver = final velocity of the second diver then
v-u = constant
That means
S or distance between them is
= (v^2-u^2)/(2×g)
= (v+u)×(v-u)÷(2×g)
S = k×(v+u) where k = constant
Thus as v and u increases the distance between the two divers S increase
