Question

Particle moves in a parabolic path defined by the vector-valued function   2 2 r t t i t j   5 where t measures in seconds. (a) find the velocity, acceleration and speed as function of time (b)sketch the curve along with the velocity vector at time =1.

Answer 1

To find velocity, acceleration, and speed as a function of time for a particle in a parabolic path, differentiate the position function with respect to time. The velocity is 4ti + j and the acceleration is 4i. The speed is given by sqrt(16t^2 + 1). To sketch the curve along with the velocity vector at t = 1, substitute t = 1 into the position function and find the corresponding point and velocity vector.

Step-by-step explanation:

To find the velocity, acceleration, and speed as a function of time, we need to differentiate the position function with respect to time. The given position function is r(t) = 2t^2i + tj -5. Let's find the velocity first:

1. Derivative of r(t) with respect to t gives us the velocity function v(t) = 4ti + j.
2. To find the acceleration, we differentiate the velocity function with respect to time. The derivative of v(t) with respect to t gives us the acceleration function a(t) = 4i.
3. The speed at any time is the magnitude of the velocity vector, which is given by the formula |v(t)| = sqrt((4t)^2 + 1^2) = sqrt(16t^2 + 1).

(b) To sketch the curve along with the velocity vector at time t = 1, we substitute t = 1 into the position function to find the point on the curve. The position at t = 1 is r(1) = 2(1)^2i + 1j -5 = 2i + j - 5. The velocity at t = 1 is v(1) = 4(1)i + j = 4i + j. So, the point on the curve at t = 1 is (2, 1 - 5) and the velocity vector is (4, 1).