A small 8.90-kg rocket burns fuel that exerts a time-varying upward force on the

Question

A small 8.90-kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F=A+Bt2. Measurements show that at t=0, the force is 121.0 N , and at the end of the first 2.30 s , it is 155.0 N .

Find the constant A.

Find the constants B.

Find the net force on this rocket the instant after the fuel ignites.

Find the acceleration of this rocket the instant after the fuel ignites

Find the net force on this rocket 3.70 s after fuel ignition.

Find the acceleration of this rocket 3.70 s after fuel ignition.

Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.70 s after fuel ignition?

Explanation / Answer

F = A + B t^2

at t =0, F = 121 N

121 = A + B (0)

A = 121 N ..........Ans

at t= 2.30 s, F = 155 N

155 = 121 + B (2.30^2)

B = 6.43 N / m^2 .......Ans

at t = 0

Fnet = F - mg = 121 - (8.9 x 9.8) = 33.78 N .........Ans

using Fnet = ma

F - mg = ma

121 - (8.90 x 9.8) = 8.9a

a = 3.8 m/s^2 .........Ans

at t = 3.70 s

Fnet = F - mg = [ 121 + 6.43(3.70^2)] - (8.9 x 9.8) = 121.81 N .....Ans

a = Fnet / m =121.08 / 8.9 = 13.7 m/s^2 ......Ans

when in space,

Fnet = F = [ 121 + 6.43(3.70^2)] = 209.03 N

a = Fnet / m = 209.03 / 8.9 = 23.5 m/s^2 ......Ans

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