Question

9. A particle of mass 1 kg moving with initial velocity
(i^+2k^) m/s is acted upon by a constant force
(i^+2j^−2k^) N. Calculate the velocity and distance after 5s.

Answer 1

Answer on Question #61860-Physics-Mechanics

A particle of mass 1 kg moving with initial velocity (i^+2k^) m/s is acted upon by a constant force (i^+2j^-2k^) N. Calculate the velocity and distance after 5s.

Solution

$$\boldsymbol {a} = \frac {\boldsymbol {F}}{m} = \frac {(1, 2, - 2)}{1} = (1, 2, - 2) \frac {m}{s ^ {2}}.$$

The velocity is

$$\boldsymbol {v} = \boldsymbol {v} _ {i} + \boldsymbol {a} t = (1, 0, 2) + 5 (1, 2, - 2) = (6, 1 0, - 8) \frac {m}{s} = (6 \boldsymbol {i} + 1 0 \boldsymbol {j} - 8 \boldsymbol {k}) \frac {m}{s}$$

The displacement is

$$\boldsymbol {s} = \boldsymbol {v} _ {i} t + \frac {\boldsymbol {a} t ^ {2}}{2} = (1, 0, 2) 5 + \frac {5 ^ {2}}{2} (1, 2, - 2) = (6, 1 0, - 8) m = (1 7. 5 \boldsymbol {i} + 2 5 \boldsymbol {j} - 1 5 \boldsymbol {k}) m$$

The distance is

$$s = \sqrt {(1 7 . 5) ^ {2} + (2 5) ^ {2} + (- 1 5) ^ {2}} = 3 4 m.$$

http://www.AssignmentExpert.com