Answer to Question #101127 in Calculus for Jason

Question

The velocity of the rocket after departure is directly proportional for some time
from the time of departure to the square root. Four seconds after departure
the rocket speed was 151.2 km / h. Calculate how high the rocket is at the time
t = 16 seconds? First, construct and integrate the rocket height equation
it. Remember that velocity v (t) = ds / dt, so that s = ∫

Answer 1

$151.2 \;km/h=151.2\frac{1000}{3600}\;m/s=42\;m/s$

$v(t)=k\sqrt{t}$

$v(4)=42\to\;2k=42\to\;k=\frac{42}{2}=21$

$s(t)=\int_0^tv(t)dt$

$s(16)=\int_0^{16}21\sqrt{t}dt=21*\frac{2}{3}t\sqrt{t}|_{t=0}^{t=16}=896\;m$