alleryStart learning →On a velocity against time graph the tangent at 25 s runs from 210 m/s at 1.0 s to 260 m/s at 51 s. What is the acceleration at 25 s?
A0.20 m/s^2
B1.0 m/s^2
C5.0 m/s^2
D50 m/s^2
Answer & Solution
Correct answer: B. 1.0 m/s^2
1. The slope of a velocity against time graph is the acceleration.
2. At a point on a curve that slope is read from the tangent line drawn there.
3. The rise is the change in velocity: 260 m/s - 210 m/s = 50 m/s.
4. The run is the change in time: 51 s - 1.0 s = 50 s.
5. Substitute the values: (50 m/s) / (50 s) = 1.0 m/s^2.
6. Quoting 50 m/s^2 reports the rise alone and never divides by the run.
7. Quoting 5.0 m/s^2 is the constant acceleration of an earlier stretch of the same journey.
_Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.8 Graphical Analysis of One-Dimensional Motion_
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