A particle is moving with the given data. Find the position of the particle.
v(t) = 1.5sqrt(t), s(9) = 33
Well, it seems like the particle is quite a mover! Let's find its position, shall we?
To do that, we need to integrate the velocity equation, v(t), with respect to time, t. But first, let's find the antiderivative of 1.5sqrt(t).
The antiderivative of 1.5sqrt(t) is 1.5 * (2/3)t^(3/2), so the position equation, s(t), would be:
s(t) = 1.5 * (2/3)t^(3/2) + C
Now, we can find the value of C by substituting in the given information, s(9) = 33.
Plugging in t = 9, we get:
33 = 1.5 * (2/3)(9)^(3/2) + C
After some mathemagics, we find that C = 33 - 9, which is 24.
Therefore, the position equation becomes:
s(t) = 1.5 * (2/3)t^(3/2) + 24
And there you have it! The position of the particle can be determined using the equation s(t) = 1.5 * (2/3)t^(3/2) + 24.
To find the position of the particle, we need to integrate the velocity function.
Given that the velocity function is v(t) = 1.5sqrt(t), we can find the position function s(t) by integrating v(t) with respect to t.
According to the Fundamental Theorem of Calculus, the position function s(t) is the integral of v(t) with respect to t and can be represented as:
s(t) = ∫(1.5sqrt(t))dt
To integrate 1.5sqrt(t), we can use the power rule for integration:
∫(t^n)dt = (t^(n+1))/(n+1)
In this case, n = 1/2 since sqrt(t) can also be written as t^(1/2).
Therefore, the integral becomes:
s(t) = ∫(1.5sqrt(t))dt = 1.5∫(t^(1/2))dt = 1.5 * (t^(1/2 + 1))/(1/2 + 1)
s(t) = 1.5 * 2(t^(3/2))/(3/2)
s(t) = 3(t^(3/2))/(3/2) = 2(t^(3/2))
Now, to find the position of the particle at t = 9, we substitute t = 9 into the position function s(t):
s(9) = 2(9^(3/2))
Calculating this expression, we get:
s(9) = 2(27)
s(9) = 54
Therefore, the position of the particle at t = 9 is 54 units.
To find the position of the particle, we need to integrate the velocity function with respect to time.
Given: v(t) = 1.5√(t)
We will use the integral of v(t) to find the position function s(t). The position function is the antiderivative of the velocity function.
To start, we integrate v(t) = 1.5√(t) as follows:
∫v(t) dt = ∫1.5√(t) dt
To integrate this expression, we can use the power rule for integration. For a term of the form ax^n, the integral is (a/(n+1))x^(n+1) + C, where C is the constant of integration.
Applying the power rule, we get:
= 1.5 * ∫√(t) dt
Now, we apply the power rule with n = 1/2:
= 1.5 * (2/3) * (√(t))^3/2 + C
= 1.5 * (2/3) * t^(3/2) + C
Since we are given that s(9) = 33, we can use this information to find the constant of integration, C.
Plugging in t = 9 and s = 33, we get:
33 = 1.5 * (2/3) * 9^(3/2) + C
33 = 1.5 * (2/3) * 27 + C
33 = 18 + C
C = 33 - 18
C = 15
Now we have the position function:
s(t) = 1.5 * (2/3) * t^(3/2) + 15
To find the position of the particle at a specific time, substitute that time into the position function. For example, to find the position at t = 9:
s(9) = 1.5 * (2/3) * 9^(3/2) + 15
s(9) = 1.5 * (2/3) * 27 + 15
s(9) = 18 + 15
s(9) = 33
Therefore, the position of the particle at t = 9 is 33.
v(t) = 3/2 t^(1/2)
s(t) = t^(3/2) + c
33 = 9^(3/2)+c, so c=6
s(t) = t^(3/2) + 6