Question

For 0 ≤ t ≤ 8 a particle moving in the xy-plane has position vector {x(t),y(t)} = {sin(2t),t²-t}, where x(t) and y(t) are measured in meters and t is measured in seconds. Find the speed of the particle at time t=2 seconds.

Answer 1

To calculate the speed of the particle at t=2 seconds, differentiate the given position vector, find the velocity components at t=2, and take the magnitude of the velocity vector. The speed is found using the square root of the sum of the squares of the velocity components.

Step-by-step explanation:

To find the speed of the particle at time t=2 seconds, we need to calculate the magnitude of the velocity vector at that instant. The velocity vector is the derivative of the position vector with respect to time. The particle's position vector is given by {x(t), y(t)} = {sin(2t), t²-t}.

First, we differentiate x(t) and y(t) with respect to time t to find the components of velocity:

• vx(t) = derivative of x(t) with respect to t = derivative of sin(2t) = 2*cos(2t)
• vy(t) = derivative of y(t) with respect to t = derivative of t²-t = 2t-1

At t = 2, the velocity components are:

• vx(2) = 2*cos(4) m/s
• vy(2) = 2*2-1 m/s = 3 m/s

Finally, the speed is the magnitude of the velocity vector and is given by:

speed = √((vx(2))² + (vy(2))²) m/s

speed at t = 2 = √((2*cos(4))² + (3)²) m/s