A particle travels along a straight line such that in 2 s it moved from an initial position Sa + 0.5 m to a position Sb -1.5 m. Then in another 4s it moves from Sb to Sc +2.5 m. Determine the particle's average velocity amd average speed during the 6-s time interval
Best Answer
avg. velocity = net displacement / timeView comments (1)
avg. velocity = +2/6 = 0.333 m/s
avg speed = total distance /time
avg speed = 2 + 4 /6 = 1 m/s
- We know that the average rate of change of a function from a to b is:View comments (1)
[f(b) - f(a)] / (b - a)
Since the average rate of change (in this case) relates to average velocity, we just plug our limits in and we get them. We see that:
f(b) = b² - 8b + 18
f(a) = a² - 8a + 18
Then:
[f(b) - f(a)] / (b - a)
= [(b² - 8b + 18) - (a² - 8a + 18)] / (b - a)
= (b² - 8b - a² + 8a) / (b - a)
= (b² - a² + 8a - 8b) / (b - a)
= [(b + a)(b - a) + 8(a - b)] / (b - a)
= [(b + a)(b - a) - 8(b - a)] / (b - a)
= b + a - 8
Then, the average velocity on [a, b] is b + a - 8.
Finally:
1) Avg velocity over [3, 4] = 3 + 4 - 8 = -1 m/s
2) Avg velocity over [3.5, 4] = 3.5 + 4 - 8 = -0.5 m/s
3) Avg velocity over [4, 5] = 4 + 5 - 8 = 1 m/s
4) Avg velocity over [4, 4.5] = 4 + 4.5 - 8 = 0.5 m/s
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