Question

25 POINTS IF YOU ANSWER BOTH

Part A First, record some data for this comparison in the table below. Record your measured values of displacement and velocity for times t = 8.0 seconds and t= 10,0 seconds in the columns below. • Next, use the measured displacement and velocity values at t=7.0 seconds and t= 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds. • Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and they values will be either displacement (x) or velocity (Vx)
y-y0=[(y1-y0)/(x1-x0) x (x-x0)
(SEE ATTACHED IMAGE)

PART B
You should find that your interpolated and extrapolated values are not even close to the actual recorded values for these displacement and velocity readings. Describe the basic assumption behind interpolation and extrapolation, then for at least one of these values explain why the calculated value was significantly larger ir smaller then the recorded value.​

Answer 1

Using kinematic principles, students estimate the displacement and velocity at certain times through interpolation or extrapolation. If calculated values differ from recorded ones, the reasons could be non-linear motion, the presence of acceleration, or measurement errors.

Step-by-step explanation:

The process described in the question involves using the principles of kinematics to calculate displacement and velocity of an object in motion at different times. When interpolating values, we use existing data points to estimate a value within the interval that we have data for, while extrapolation involves estimating values outside the interval of known data points.

The formula provided, y-y0=[(y1-y0)/(x1-x0) x (x-x0)], is used to find the y-value (in this case, displacement or velocity) at a specific x-value (time). The basic assumption behind these methods is that the change between values is linear and can be modeled by a straight line. If a student finds that the calculated values are significantly different from the actual recorded values, it could be due to the actual motion not being uniformly linear, the presence of acceleration or deceleration, or measurement errors.