A time-dependent force, $ F = \left( {8.00i{\text{ }}-{\text{ }}4.00tj} \right)N $ , where $ t $ is in seconds, is exerted on a $ 2.00kg $ object initially at rest.
A) At what time will the object be moving with a speed of $ 15.0{\text{ }}m/s $ ?
B) How far is the object from its initial position when its speed is $ 15.0{\text{ }}m/s $ ?
C) Through what total displacement has the object travelled at this moment?

Hint: For part A) first find the acceleration using Newton's second law. Integrate it with respect to time. We will get a relation between velocity and time. Find the value of time. For part B) use the formula of distance to find the distance of the object from its initial position. for part C) find the displacement by the difference between final and initial position of the object.

Complete Step By Step Answer:
It is given that the $ F = \left( {8.00i{\text{ }}-{\text{ }}4.00tj} \right)N $ mass of the object is $ 2.00kg $ and speed of the object is $ 15.0{\text{ }}m/s $

Solution of part A)
Using Newton's second law of motion.
 $ F = ma $
 $ \Rightarrow a = \dfrac{F}{m} $
 $ \Rightarrow a = \dfrac{{8.00i + (4.00t)j}}{2} $
 $ \Rightarrow a = 4.00i + (2.00t)j $
Integrating the above equation
 $ \dfrac{{dv}}{{dt}} = 4.00i + (2.00t)j $
 $ \Rightarrow v = (4.00t)i + ({t^2})j $
Putting the value of speed
 $ \Rightarrow 15 = (4.00t)i + ({t^2})j $
 $ \Rightarrow {15^2} = {(4.00t)^2} + {({t^2})^2} $
 $ \Rightarrow {t^4} + 16{t^2} = 225 $
 $ \Rightarrow {t^2} = 9\sec $
 $ \Rightarrow t = 3\sec $
Hence, the time in which the object will be moving with a speed of $ 15.0{\text{ }}m/s $ is $ 3\sec $

Solution for part B)
Using the distance formula speed is equal to distance divided by time
 $ \dfrac{{dx}}{{dt}} = (4.00t)i + ({t^2})j $
On integrating the above equation
 $ x = (2.00{t^2})i + \dfrac{{{t^3}}}{3}j $
Putting the value of time from above,
 $ \left| x \right| = \sqrt {(2 \times {3^2}) + \dfrac{{(3){}^3}}{3}} $
 $ \Rightarrow x = 5.19m $
Hence, the object is distance from its initial position when its speed is $ 15.0{\text{ }}m/s $ is $ 5.19m $

Solution for part C)
the total displacement the object has travelled at this moment in vector form is
 $ d = 18i - 9j $ .

Note:
The three Newtonian laws of motion give the relation between an object's motion and the forces acting on it. According to Newton's second law, the rate of change of momentum is proportional to the force applied for the given object, or that the force on an object is equal to the product of mass and the acceleration.