1. Given condition

A force of constant magnitude acts on the particle, and it is always perpendicular to the velocity.

So at every instant, $\vec F \perp \vec v$

2. Effect on kinetic energy

The rate of work done by force is $\frac{dW}{dt}=\vec F\cdot \vec v$

Since the force is always perpendicular to velocity, $\vec F\cdot \vec v = Fv\cos 90^\circ = 0$

Hence, $\frac{dW}{dt}=0$

By work-energy theorem, $\frac{d}{dt}\left(\frac12 mv^2\right)=0$

Therefore, $\frac12 mv^2=\text{constant}$

So the kinetic energy remains constant.

3. Check other options

• Option B: acceleration is constant

  Acceleration is $\vec a=\frac{\vec F}{m}$ Magnitude of acceleration is constant because $F$ is constant, but its direction changes continuously since it must always remain perpendicular to velocity. Therefore, acceleration vector is not constant.

  So B is false.

• Option C: velocity is constant

  Constant kinetic energy means constant speed, not constant velocity. Since force is perpendicular to velocity, it changes the direction of velocity.

  So C is false.

• Option D: it moves in a straight line

  If velocity direction changes, motion cannot be in a straight line. In fact, such motion is curved (for example, uniform circular motion is a standard case).

  So D is false.

4. Conclusion

The correct option is: $\boxed{A}$

The particle's kinetic energy is constant.