A car drives down a road in such a way that its velocity (in m/s) at time t (seconds) is v(t)=2t^(1/2)+3.
Find the car's average velocity (in m/s) between t=5 and t=8.
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So I used integrals for this and put ∫2t^(1/2)+3 and got 4/3(t^(3/2))+3t.
Then I substituted in 8 and 5 into that and subtracted that, but it's still marked wrong? I even entered it into Symbolab to check but that was also incorrect.
(I got ((64√2 - 20√5)/3)+9 from symbolab)
average value of f(x) over the interval [a,b] is
∫[a..b] f(x) dx
----------------------
(b-a)
You did the integral, but did not divide by 3
To find the car's average velocity between t=5 and t=8, you need to use the definite integral of the velocity function over the given time interval.
First, let's review the steps to find the average velocity:
1. Calculate the integral of the velocity function v(t) with respect to t.
∫(2t^(1/2) + 3) dt
2. Evaluate the definite integral over the given time interval [5, 8].
∫(2t^(1/2) + 3) dt evaluated from 5 to 8
3. Evaluate the integral at the upper limit (8) and subtract it from the evaluation at the lower limit (5).
[∫(2t^(1/2) + 3) dt] from 5 to 8
= [4/3(t^(3/2)) + 3t] evaluated from 5 to 8
= [4/3(8^(3/2)) + 3(8)] - [4/3(5^(3/2)) + 3(5)]
Let's calculate this step by step:
1. Calculate the integral of the velocity function:
∫(2t^(1/2) + 3) dt = [4/3(t^(3/2)) + 3t] + C, where C is the constant of integration.
2. Evaluate the definite integral over the given time interval:
[4/3(t^(3/2)) + 3t] evaluated from 5 to 8
= [4/3(8^(3/2)) + 3(8)] - [4/3(5^(3/2)) + 3(5)]
Now, let's calculate the result:
[4/3(8^(3/2)) + 3(8)] - [4/3(5^(3/2)) + 3(5)]
≈ (155.49 + 24) - (44.72 + 15)
≈ 179.49 - 59.72
≈ 119.77 m/s
Therefore, the car's average velocity between t=5 and t=8 is approximately 119.77 m/s.