3. What are the velocity and acceleration function vectors for a particle traveling in the xy-plane with position vector r(t)=(2+t^2, 3t)?
a)v(t)=(2t^2, 3t) ; a(t)=(2t, 3)
b)v(t)=(2t, 3) ; a(t)=(2, 0)
c)v(t)=(2t, 3t) ; a(t)=(2, 3t)
d)v(t)=(2+2t, 3) ; a(t)=(4, 0)
Hey, now I showed you how to do two of them. In fact this looks very much like one I just did.
To find the velocity and acceleration function vectors for the particle traveling in the xy-plane with position vector r(t) = (2 + t^2, 3t), we can differentiate the position vector with respect to time.
First, let's find the velocity vector v(t):
v(t) = dr(t)/dt
Differentiating each component of r(t) with respect to t, we get:
v(t) = (d/dt)(2 + t^2, 3t)
= (d/dt)(2 + t^2, 3t)
= (2t, 3)
So the velocity function vector is v(t) = (2t, 3).
Next, let's find the acceleration vector a(t):
a(t) = dv(t)/dt
Differentiating each component of v(t) with respect to t, we get:
a(t) = (d/dt)(2t, 3)
= (2, 0)
So the acceleration function vector is a(t) = (2, 0).
Comparing the given options with the calculated velocity and acceleration function vectors, we see that the correct answer is b) v(t) = (2t, 3) and a(t) = (2, 0).