If you dive straight down into a pool of water with a speed of 5.0 m/s, you are immediately subject to a drag force of fd = −(1.00 × 104 kg/s)v, pointing opposite your velocity. if your mass is 70 kg, how long does it take you to lose 98% of your original speed?

 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

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y