Question

For 0 515 8, a particle moving in the xy-plane has position vector (x(t), y(t)) = (sin(2), t² – t), where x(t) and y(t) are measured in meters and t is measured in seconds.

(a) Find the speed of the particle at time t = 2 seconds. Indicate units of measure.

Answer 1

The speed of the particle at time t = 2 seconds is 3.1546 m/s.

Step-by-step explanation:

To find the speed of the particle at time t = 2 seconds, we need to find the magnitude of the velocity vector at t = 2 seconds. The velocity vector is given by v(t) = (x'(t), y'(t)), where x'(t) and y'(t) are the derivatives of x(t) and y(t) with respect to t.

Given x(t) = sin(2t) and y(t)

= t^2 - t, we can find the derivatives:

x'(t) = cos(2t) * 2 and

y'(t) = 2t - 1.

Substituting t = 2 into x'(t) and y'(t), we get:

x'(2) = cos(4) * 2 and

y'(2) = 2(2) - 1.

Calculating the values:

x'(2) = -0.6536 m/s and y'(2)

= 3 m/s.

The magnitude of the velocity vector is given by:

speed = sqrt((x'(2))^2 + (y'(2))^2).

Substituting the values we calculated:

speed = sqrt((-0.6536)^2 + (3)^2)

= 3.1546 m/s.