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The force F on an object is the product of the mass m and the acceleration a. In this problem, assume that the mass and acceleration both depend on time t, hence so the does the force. That is,F(t) = m(t)a(t)At time t=10 seconds, the mass of an object is 47 grams and changing at a rate of -6 \frac{g}{s}. At this same time, the acceleration is 17 \frac{m}{s^2} and changing at a rate of -7 \frac{m}{s^3}.By the product rule, the force on the object is changing at the rate of

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Answer:

D(F)/ dt =  - 431  grs*m/s³

Step-by-step explanation:

F(t)  =  M(t)*a(t)

Taking derivatives on both sides of the equation we get

D(F)/ dt  =  DM(t)/dt * a(t)  + Da(t)/dt* M(t)      (1)

At time t = 10 s

M(t) = M(10) = 47 grs     and     DM(t)/dt = -6 grs/s

a(t)  =  a (10) = 17 m/s²   and      Da(t)/dt = -7 m/ s³

Plugging these values in equation (1) we get

D(F)/ dt  =  DM(t)/dt * a(t)  + Da(t)/dt* M(t)

D(F)/ dt = - 6 grs/s * 17 m/s²  + (-7) m/s³ *47 grs

D(F)/ dt =  -  102 grs*m/s³  - 329 grs*m/s³

D(F)/ dt =  - 431  grs*m/s³

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