A particle moves at a speed of 2.5 m/s in the +x direction. Upon reaching the origin, the particle receives?

Question

a continuous constant acceleration of .75 m/s squared in the -y direction. What is the position of the particle 4.0 seconds later? Thank you

Answer 1

Keep in mind that the motion in the x direction is independent of the motion in the y direction so you can work those separately.

Start with x. Particle's initial position is 0 and experiences no acceleration in x. So at a time t later it is at:

x = vx*t ---> x = 2.5*4 = 10 m

Now in the y direction, the particle starts from 0, has no initial speed in the y direction and has a constant acceleration a in the -y direction, so it's position in y is given by:

y = 1/2at^2 = 1/2*(-0.75 m/s^2)*(4 sec)^2 = -6 m

So after four seconds the particle is at (10, -6) m