Answer:
t = 0.0275 s
Explanation:
Let's use Newton's second law
W - fr = m a
m g - 1 10⁴ v = m dv / dt
dt = dv / (g-1 10⁴/m v)
We change variables
u = g- 1 10⁴ /m v
du = - 1 10⁴ / m dv
dt = -m / 1 10⁴ du / u
We integrate
t = -m / 1 10⁴ ln u
t = -m / 1 10⁴ ln (g- 1 10⁴ /m v)
For t = 0 v = v₀ and for a time t the speed is v
t- 0 = - m / 1 10⁴ [ln (g-10 10⁴ v) - ln (g- 1 10⁴ vo)]
- t 1 10⁴ / m = ln ((g-1 10⁴ v) / (g-1 10⁴ v₀))
((g-1 10⁴ v) / (g-1 10⁴ v₀)) = e ( - 1 10⁴ /m t)
Let's replace and calculate
(9.8 - 1 10⁴ v) / (9.8 -1 10⁴ 5.0) = e ( -1 10⁴/70 t)
(9.8-1 10⁴ v) / (-49990.2) = e (- 142.86 t)
They ask us for the time for when they have lost 98% of the initial speed, so that he has a 2% vo
v = 0.02 vo
v = 0.02 5 = 0.1 m / s
We substitute and simplify
(9.8 - 1 10⁴ 0.1) (-49990.2) = e ( -142.86 t)
0.01981 = e ( -142.86 t)
- 142.86 t = ln (0.01981)
. t = 0.0275 s
