CP CALC? A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by v(t) = At + Bt2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Solution 22E Step 1 of 6: (a) Determine A and B , including their SI units. Given vertical velocity as the function of time, v = At + Bt ……….1 Solving for A, Differentiating with respect to time, d d 2 dtv = dt(At + Bt ) Using a= v d a = d(At) + (Bt ) 2 dt dt dt a= A+ 2Bt 2 Substituting at t= 0s , a= 1.5m/s 1.5m/s = A+ 0 2 Therefore, A= 1.5m/s Step 2 of 6: Solving for B Substituting A= 1.5m/s ,v=2 m/s at t=1s in equation 1, 2 v = At + Bt 2 m/s=1.5m/s (1s)+ B(1s) 2 2 B (1s) = 2 m/s -1.5 m/s B (1s) = 0.5 m/s B =0.5 m/s 3 3 2 Using B =0.25 m/s and A= 1.5m/s in equation 1 v = (1.5m/s )t + (0.5 m/s )t ………...2 Similarly acceleration equation becomes, a= (1.5m/s )+ 2(0.5 m/s )t 3 a= 1.5+ t…………..3 Step 3 of 6: (b) t 4.00 s after fuel ignition, what is the acceleration of the rocket, In order to calculate the acceleration at t=4s, we need to calculate the velocity at the same instant. Substituting t = 4s in equation 2, v = (1.5m/s )(4 s) + (0.5 m/s )(4 s) 2 V = 8 m/s By fundamental relation of acceleration, a= v t Using V = 7 m/s and t= 4s 8 m/s a = 4 s 2 a= 2 m/s 2 Therefore, acceleration 4s after the fuel ignition is 2 m/s .