Question

At t = 0, a particle starts from rest at x = 0, y = 0, and moves in the xy plane with an acceleration aƒ— = (4.0i^ + 3.0j^) m/s². Assume t is in seconds. What is the position of the particle at t = 2 seconds?

Answer 1

The position of the particle at t = 2 seconds is 8.0i^ + 6.0j^ + 2C1 + C2.

Step-by-step explanation:

The position of the particle can be found by integrating the acceleration twice with respect to time. Let's start by finding the velocity function. Since the particle starts from rest, we know that the initial velocity is zero. The velocity function can be found by integrating the acceleration function:

v(t) = ∫(aƒ—) dt = ∫(4.0i^ + 3.0j^) dt = (4.0t)i^ + (3.0t)j^ + C1

Next, we can find the position function by integrating the velocity function:

r(t) = ∫(v(t)) dt = ∫((4.0t)i^ + (3.0t)j^ + C1) dt = (2.0t^2)i^ + (1.5t^2)j^ + C1t + C2

Now, we can substitute t = 2 seconds into the position function to find the position of the particle:

r(2) = (2.0(2)^2)i^ + (1.5(2)^2)j^ + C1(2) + C2

r(2) = 8.0i^ + 6.0j^ + 2C1 + C2