Around midcourse, a velocity adjustment is performed to eliminate the small errors introduced when departing from earth's orbit. this adjustment is performed using one of the onboard thrusters. at the location where the adjustment is made, the velocity is 26533 m/s and should be 27000 m/s. knowing that the thruster used for the maneuver generates a thrust of f= 7880 n, determine how long, in minutes, it should be turned on to adjust the velocity. the mass of the spacecraft is 2000 kg.

To solve this complicated, perplexing problem, we must
need a complicated, perplexing formula, right ? 

Sure.  Here's the formula we need, and I'll even bet that
you've seen it before:

                         F  =  M · a

                       Force = (mass) x (acceleration)

We know the force of the thruster, and we know the mass
of the spacecraft, so we can calculate the acceleration of
the spacecraft whenever the thruster is thrusting:

                              7,880 Newtons = (2,000 kg) x (acceleration)

Divide each side
by  2,000 kg:          (7,880 kg-m/s²) / (2,000 kg)  =  acceleration

                                   3.94  m/s²  =  acceleration with thruster thrusting.

Now, we need to change the spacecraft's speed from 26,533 m/s
to 27,000 m/s. 

What's the change ?  It's  (27,000 - 26,533)  =  467 m/s faster.

We need 467 m/s more speed, and we have a thruster that can give us
3.94 more m/s of speed for every second that we keep it thrusting.
How many seconds should we let it keep thrusting ?

                       (467 m/s) / (3.94 m/s²)  =  118.5 seconds .

Sadly, ground control can't understand anything smaller than minutes,
so that's the way we'll need to report it to them.

          (118.5 seconds) x (1 minute / 60 seconds)  =  1.975 minutes .    

It is d, all the other ones say incorrect answer .