On a curved position against time graph, the tangent drawn at 25 s runs from a position of 1300 m at 19 s to a position of 3120 m at 32 s. What is the instantaneous velocity at 25 s?

1. The slope of a curve at a point equals the slope of the straight line tangent at that point. 2. The instantaneous velocity at 25 s is therefore the slope of that tangent. 3. The rise is the change in position: 3120 m - 1300 m = 1820 m. 4. The run is the change in time: 32 s - 19 s = 13 s. 5. Substitute the values: (1820 m) / (13 s) = 140 m/s. 6. Using the whole curve rather than the tangent would give an average velocity, not the value at 25 s. 7. Halving the rise before dividing gives 70 m/s, which is the arithmetic trap. _Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.8 Graphical Analysis of One-Dimensional Motion_