Question

The position of a particle in the​ xy-plane at time t is r(t)=(t−5) i+t2+4 j. Find an equation in x and y whose graph is the path of the particle. Then find the​ particle's velocity and acceleration vectors at t=5.

Answer 1

The equation in x and y whose graph is the path of the particle is y = x^2+4. The particle's velocity vector at t=5 is v(5) = i + 10j. The particle's acceleration vector at t=5 is a(5) = 2j.

Step-by-step explanation:

To find an equation in x and y whose graph is the path of the particle, we can rewrite the position vector r(t) = (t−5)i + t^2+4j as x = t−5 and y = t^2+4. Therefore, the equation in x and y is y = x^2+4.

To find the particle's velocity vector, we can take the derivative of the position vector with respect to time. The velocity vector is given by v(t) = d/dt [(t−5)i + t^2+4j] = i + 2tj. At t = 5, the velocity vector is v(5) = i + 2(5)j = i + 10j.

To find the particle's acceleration vector, we can take the derivative of the velocity vector with respect to time. The acceleration vector is given by a(t) = d/dt [i + 2tj] = 0i + 2j. At t = 5, the acceleration vector is a(5) = 0i + 2j = 2j.