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Average velocity formula calculus9/23/2023 ![]() ![]() ![]() What is his average velocity?Ī runner is running around rectangle track with length = 50 meters and width = 20 meters. He then reverses and drives 12 km back down the road in 3 minutes. Determine average velocity.Ī truck driver drives 20 km down the road in 5 minutes. The average velocity between t = 1 and t = 2 is given byĪ car travels along a straight road to the east for 120 meters in 4 seconds, then go to the west for 40 meters in 1 second. What is the average velocity between t = 1 and t = 2 seconds? Where t is measured in seconds (we are neglecting air resistance). The average velocity between t = 0 and t = 10 is given byĪn object is dropped from the observation deck of the CN tower so that its height in meters is given by The position of a car is given by s = 10 + 5t + 20t 2 meters at t seconds. What is the average velocity between t = 0 and t = 10 seconds? The average velocity between t = 2 and t = 5 is given by Since we have two velocities $v$ and $v_0$, we expect the representative $\bar v$ must be contributed equally by them.Let s 1 be the position of an object at time t 1 and s 2 be the position of the same object time t 2, then the average velocity over the time interval (t 2 - t 1 ) is defined byĭisplacement : Distance between the starting and ending point.Īn object moves along a straight line so that its position in meters is given by s(t) = t 3 - 6t 2 + 9t for all time in t seconds. Find the average velocity of the object between t = 2 and t = 5 seconds. The average $\bar v$, which is some kind of representative of the initial and final velocities, thus must obey the following:Īs if $v_0=\bar v$ or $v = \bar v$, then it does not represent that there's a contribution of $v$ (respectively $v_0$) in it Now consider acceleration, and the initial velocity $v_0$, Then the final velocity $v$ must be greater than $v_0$ It just seems to pop out of nowhere without any rigorous proof.įirst, consider constant velocity $v$, then by daily intuition, the average velocity $\bar v=v$ (because its the same $v$ throughout, thus the average cannot be anything other than $v$) Is there any other way to demonstrate to a non-calculus student that it is true? I've never seen any textbook attempt to justify this formula. The area under the curve bit of knowledge implies a knowledge of calculus. Drawing a simple v/t graph that is linear (constant acceleration) I could show geometrically that the area under this curve is the same formula.Īnd there's the problem. In particular, I would use the knowledge that the area under the v/t curve is the displacement. The only problem is that the only way I can see to proving that this formula is indeed true is to use calculus! That the average velocity is the mid-point between the initial and final velocities makes intuitive sense I guess, but I want to be able to demonstrate to students that this is true in a more rigorous way. But if you don't have a calculus background I guess using this first equation is a necessary stepping stone on the way to deriving the formula. We just keep integrating from a = constant until we get the same formula. With a calculus background we don't even need this formula. This formula is often substituted into the $x = x_0 + \overline v t$ formula in order to derive the $x = x_0 + v_0t + 1/2at^2$ formula we are so familiar with. In physics textbooks, to describe the average velocity for an object moving with constant acceleration. ![]()
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