Question

Part A: Resolve t = 8 (interpolate)

Answer 1

The displacement and velocity values at time, t = 8 seconds are; 11.76 and -0.545 respectively

How can interpolating be used to find variable values at a given time?

Part A

The possible values in the table are;

Displacement;

t = 7 seconds, y = 10.02

t = 8 seconds, y = 11.01

t = 9 seconds, y = 13.5

Velocity;

t = 7 seconds, y = 0.06

t = 8 seconds, y = -0.05

t = 9 seconds, y = -1.15

The equation for interpretation which is a form of linear equation is presented as follows;

$y - y_0 = \frac{y_1 - y_0}{x_1 -x_0} \times (x - x_0)$

Interpolating for the displacement at t = 8.0 seconds is as follows;

$y - 10.02 = \frac{13.5 - 10.02}{9 - 7} \times (8 - 7)$

Which gives the displacement at t = 8.0 seconds as 11.76

Interpolating for the velocity at t = 8.0 seconds is as follows;

$y - 0.06 = \frac{(-1.15) - 0.06}{9 - 7} \times (8 - 7)$

Which gives the velocity at t = 8.0 seconds as -0.545

The values for the displacement and velocity obtained by interpolation (11.76 and -0.545) are different from the displacement and velocity values in the table at t = 8 which are; 11.01 and -0.05 respectively

Learn more about linear equations here:

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