My physics textbook claimed that if velocity was constant over an interval, then the average velocity was that constant velocity. The book used this to prove that the area under a curve with constant velocity is the displacement(since average velocity = displacement/time). My question is how could you (only using the definition of average velocity) initially say that average velocity is equal to the constant velocity? Wouldn't you need to first know that displacement was the area under the velocity curve?
2 Answers
An average is always between the minimum and maximum. So, the average velocity is between the minimum velocity and the maximum velocity. Since the velocity is constant, the minimum and maximum velocities equal this constant, so the average velocity must be that constant.
In symbols, $$c = v_{\text{min}} \le v_{\text{average}} \le v_{\text{max}} = c,$$ which implies that $v_{\text{average}} = c$.
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Average velocity $= \Delta s/\Delta t$. Total distance over total time.
Constant instantaneous velocity $= ds/dt = v_0$. Instantaneous change in position over instantaneous change in time.
$\Delta s=\int_{t_1}^{t_2}\frac{ds}{dt} dt=\int_{t_1}^{t_2} v_0 dt=v_0(t_2-t_1)=v_0\Delta t$. Total displacement is the integral of velocity with respect to time.
So $\Delta s/\Delta t=v_0$
So constant instantaneous velocity implies the average velocity is equal to that velocity.
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