The only force acting on a 3.1 kg body as it moves along the positive x axis has an x component fx = -8x n, where x is in meters. the velocity of the body at x = 2.0 m is 11 m/s. (a) what is the velocity of the body at x = 4.5 m? (b) at what positive value of x will the body have a velocity of 5.8 m/s?

(a) -9.97 m/s
 (b) x = 2.83

   This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.

   f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2

   So the acceleration of the body is now expressed as
 f''(x) = -2.580645161x m/s^2

    Let's calculate the anti-derivative from that.
 f''(x) = -2.580645161x m/s^2
 f'(x) = -1.290322581x^2 + C m/s

   Now let's use the known velocity value at x = 2.0 to calculate C
 f'(x) = -1.290322581x^2 + C 1
1 = -1.290322581*2^2 + C
 11 = -1.290322581*4 + C
 11 = -5.161290323 + C
 16.161290323 = C

   So the velocity function is
  f'(x) = -1.290322581x^2 + 16.161290323

   (a) The velocity at x = 4.5
  f'(x) = -1.290322581x^2 + 16.161290323
 f'(4.5) = -1.290322581*4.5^2 + 16.161290323
 f'(4.5) = -1.290322581*20.25 + 16.161290323
  f'(4.5) = -26.12903227 + 16.161290323
 f'(4.5) = -9.967741942

   So the velocity is -9.97 m/s

   (b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
 0 = -1.290322581x^2 + 10.36129032
 1.290322581x^2 = 10.36129032
 x^2 = 8.029999998
 x = 2.833725463

Learned about life itself