Suppose that the acceleration of a model rocket is proportional to the difference between 160 ft/sec and the rocket's velocity. If it is initially at rest and its initial acceleration is 280 ft/sec22, how long will it take to accelerate to 128 ft/s

Answer :

Answer:

Explanation:

Given ,

dv / dt = k ( 160 - v )

dv / ( 160 - v ) = kdt

ln ( 160 - v ) = kt + c , where c is a constant

when t = 0 , v = 0

Putting the values , we have

c = ln 160

ln ( 160 - v ) = kt + ln 160

ln ( 160 - v / 160 ) = kt

(160 - v ) / 160 = [tex]e^{kt}[/tex]

1 - v / 160 = [tex]e^{kt }[/tex]

v / 160 = 1 - [tex]e^{kt }[/tex]

v = 160 ( 1 - [tex]e^{kt }[/tex] )

differentiating ,

dv / dt = - 160k [tex]e^{kt }[/tex]

acceleration a   = - 160k [tex]e^{kt }[/tex]

given when t = 0 , a = 280

280 = - 160 k

k = - 175

a = - 160 x - 175 [tex]e^{kt}[/tex]

a = 28000 [tex]e^{kt}[/tex]

when a = 128  t = ?

128 = 28000 [tex]e^{kt}[/tex]

[tex]e^{kt }[/tex] = .00457

Acceleration of a rocket model.

As acceleration is the change in the magnitude and direction of the moving body and refers to the increase in the velocity which varies over time. The acceleration of the model rocket is proportional to the r difference between the 160 feet/sec.

Thus the answer is 0.0457

  • The initial velocity of the rockset at the rest and the initial acceleration is about 280 fett/ sec then the length of acceleration for the rokst to reach 128 feet / sec. Will be calculated by the dv / dt = k ( 160 - v ).
  • ln ( 160 - v ) = kt + c , where c is a constant, when t = 0 , v = 0 then putting the values , we have c = ln 160
  • ln ( 160 - v ) = kt + ln 160, ln ( 160 - v / 160 ) = kt
  • (160 - v ) / 160 = 1 - v / 160 = v / 160 = 1 - v = 160 ( 1 -  ) acceleration a   = - 160k  given when t = 0 , a = 280 The 280 = - 160 k, k = - 175 a = - 160 x - 175  Hence a = 2800
  • Thus 128 = 28000  = .00457.

Learn more about acceleration.

brainly.com/question/13137118.