Average Speed = (Total distance)/(Total Time)
-secret_angel-
Answer:
V(avg) = [(v+u)(v-u)/2] ÷ (v-u) So, V(avg) = (v+u)/2
Explanation:
Let initial velocity is u and final velocity is v. Velocity is changing uniformly.
This means that the motion is uniformly accelerated.
Let acceleration be f.
Also, let distance traveled be s.
So, v² = u² + 2fs So, s = (v² - u²)/2f -----(1) Also, v = u + ft So, t = (v-u)/a -----(2)
Now, Let average speed be V(avg).
Then V(avg) = Total Distance/Total Time So, V(avg) = s/t So, V(avg) = [(v²-u²)/2a] ÷ [(v-u)/a]
From equation (1) and (2)
So, V(avg) = [(v+u)(v-u)/2] ÷ (v-u) So, V(avg) = (v+u)/2.
Thus, the average speed is the average of initial and final velocities in this case.
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