A rocket is launched from rest with a time-dependent acceleration a(t) = (p – rt)?
A rocket is launched from rest with a time-dependent acceleration a(t) = (p – rt), where p = 35.00 m/s^2 and r = 4.50 m/s^3. Find the maximum velocity of the rocket.
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dv/dt = p - rt
dv/dt = 0 when p = rt, i.e. when t = (35/4.5) s = 70/9 seconds
dv = (p - rt) dt
v = pt - (1/2) rt^2 + C, but C=0 since the rocket is launched from rest.
Thus, at t = 70/9 seconds, we have
v = (35*70/9) m/s - (1/2)(4.5*70*70/81) m/s = about 136 m/s
v(t) = integral[ a(t) dt] = pt -rt^2/2 when v = 0 at t=0
v(t) will have an extreme value when its derivative, a(t) = 0 so p - rt = 0 and t = p/r
put this time into the first eqtn. for v(t) and do the math to get your answer