What is the magnitude of the velocity vector at time t = 8.3 s?

[SMS]

A particle moves in the x-y plane with constant acceleration. At time t = 0 s, the position vector for the particle is d = 2.2 m x + 8.2 m y. The acceleration is given by the vector a = 4.5 m/s2 x + 2.2 m/s2 y. The velocity vector at time t = 0 s is v = 3.3 m/s x - 4.3 m/s y. Find the magnitude of the velocity vector at time t = 8.3 s.

I never liked vectors

 

[Galileo]

I never liked vectors

No man. Change that attitude. Vectors are like, totally awesome
You'll appreciate them when you take linear algebra.

Anyway, simply view this problem as two one-dimensional problems.
The x- and y coordinates are independent.
Since the acceleration is constant, the formula for the speed as a function of time is:
$$v_x(t)=v_{x0}+a_xt$$
And a similar equation for $v_y(t)$.

You can calculate
$$v(t)=\sqrt{v_x(t)^2+v_y^2(t)}$$
with this.

 

[arildno]

No man. Change that attitude. Vectors are like, totally awesome
You'll appreciate them when you take linear algebra.

No, they start rocking when taking physics

 

[SMS]

U guys are right

Vectors don"t suck. I think I was just frustrated with the rest of my work.

I am taking physics and that was a physics problem.

Any ways I got all the answers right on CAPA, so everything rocks!

Time to :zzz: :zzz: