Question

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity when s=2km and the time needed to reach this altitude. initially, v=0 and s=0 when t=0

Answer 1

a) velocity v = 322.5m/s

b) time t = 19.27s

Step-by-step explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

`\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + (0.02s^(2) )/(2) = (1)/(2) v^(2) \\v = \sqrt{12s + 0.02s^(2) } .....................1 \\\\\\`

Remember that

`v = (ds)/(dt) \\(ds)/(v) = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^(2) } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5√(2) ) ln \frac{| [s + 300 + \sqrt{(s^(2) + 600s)} ] |}{300} .......2`

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s