Final answer:
To find the parametric equations for the tangent line to the curve at t=25s, we need to find the derivative of the curve at that point. We can then use the point-slope form of a line equation to parametrize the line. The tangent line passes through the point (1300, 19) and has velocity components v_x = 820/13 and v_y = 13/182.
Step-by-step explanation:
The tangent line to a curve at a given point can be found by finding the derivative of the curve at that point. The derivative represents the slope of the tangent line. In this case, we need to find the parametric equations for a line tangent to the curve at t=25s.
To parametrize the line, we can use the point-slope form of a line equation. The equation of the tangent line passing through the point (1300, 19) can be written as:
x = 1300 + (t-19)v_x
y = 19 + (t-19)v_y
where v_x and v_y are the components of the velocity vector at t=19s.
To determine the values of v_x and v_y, we can use the given endpoints of the tangent line at t=19s and t=32s. The position vector at t=19s is (1300, 19), and the position vector at t=32s is (3120, 32). We can find v_x and v_y by subtracting the position vectors:
v_x = (3120-1300)/(32-19) = 820/13
v_y = (32-19)/(3120-1300) = 13/182
Therefore, the parametric equations for the tangent line are:
x = 1300 + (t-19)(820/13)
y = 19 + (t-19)(13/182)