The velocity function is v(t)=t^2-5t+6 for a particle moving along a line. Find the displacement of the particle during the time interval [-3,6].
Well, let's first find the antiderivative of the velocity function to get the position function. The antiderivative of t^2 is (1/3)t^3, the antiderivative of -5t is -2.5t^2, and the antiderivative of 6 is 6t. So the position function is given by p(t) = (1/3)t^3 - 2.5t^2 + 6t.
To find the displacement during the time interval [-3,6], we need to evaluate p(6) - p(-3). Plugging in 6 into the position function, we get p(6) = (1/3)(6)^3 - 2.5(6)^2 + 6(6). Simplifying that expression gives us p(6) = 72.
Now plugging in -3 into the position function, we get p(-3) = (1/3)(-3)^3 - 2.5(-3)^2 + 6(-3). Simplifying that expression gives us p(-3) = 51.
Finally, to find the displacement, we subtract p(-3) from p(6). So the displacement during the time interval [-3,6] is p(6) - p(-3) = 72 - 51 = 21.
Therefore, the displacement of the particle during the time interval [-3,6] is 21 units. And by the way, I hope that calculating those values didn't displace your sense of humor!
To find the displacement of the particle during the time interval [-3, 6], we need to integrate the velocity function over this interval.
The displacement of an object is given by the definite integral of its velocity function.
Given that the velocity function is v(t) = t^2 - 5t + 6, we can find the displacement as follows:
∫(t^2 - 5t + 6) dt
We can rewrite the velocity function as follows:
∫(t^2 - 5t + 6) dt = ∫t^2 dt - ∫5t dt + ∫6 dt
Now, we can integrate each term separately:
∫t^2 dt = (1/3)t^3 + C1, where C1 is the constant of integration.
∫5t dt = (5/2)t^2 + C2, where C2 is the constant of integration.
∫6 dt = 6t + C3, where C3 is the constant of integration.
Now, we can use these results to find the displacement over the time interval [-3, 6]:
Displacement = ∫(t^2 - 5t + 6) dt = [(1/3)t^3 + C1] - [(5/2)t^2 + C2] + [6t + C3]
Substituting the limits of integration, we have:
Displacement = (1/3)(6)^3 + C1 - (5/2)(6)^2 - C2 + 6(6) + C3 - [(1/3)(-3)^3 + C1 - (5/2)(-3)^2 - C2 + 6(-3) + C3]
Simplifying, we get:
Displacement = (72/3) + C1 - (180/2) - C2 + 36 + C3 - (9/3) - C1 + (45/2) + C2 - 18 + C3
C1 and C2 cancel each other out, and C3 cancels with -C3:
Displacement = 24 - 90 + 36 - 3 + 22
Displacement = -11
Therefore, the displacement of the particle during the time interval [-3, 6] is -11 units.
To find the displacement of the particle during the time interval [-3,6], you need to integrate the velocity function over that interval. The displacement is given by the definite integral of the velocity function.
The velocity function is v(t) = t^2 - 5t + 6.
To integrate this function, you need to find its antiderivative or integral.
∫(t^2 - 5t + 6) dt = ∫t^2 dt - ∫5t dt + ∫6 dt.
To integrate t^2 with respect to t, use the power rule of integration, which states that the integral of t^n is (1/(n+1)) * t^(n+1). Applying this rule, we have:
∫t^2 dt = (1/3) * t^3 + C1,
where C1 is the constant of integration.
To integrate -5t with respect to t, use the power rule again:
∫-5t dt = (-5/2) * t^2 + C2,
where C2 is another constant of integration.
Finally, integrating 6 with respect to t is straightforward:
∫6 dt = 6t + C3,
where C3 is a third constant of integration.
Putting it all together, the antiderivative of the velocity function is:
∫(t^2 - 5t + 6) dt = (1/3) * t^3 - (5/2) * t^2 + 6t + C,
where C = C1 + C2 + C3 is the combined constant of integration.
Now we can find the displacement of the particle by evaluating the definite integral of the velocity function over the given time interval:
Displacement = ∫[a,b] (t^2 - 5t + 6) dt,
where a = -3 and b = 6.
Substituting the limits of integration, we get:
Displacement = ∫[-3,6] (t^2 - 5t + 6) dt
= [(1/3) * t^3 - (5/2) * t^2 + 6t] [from -3 to 6]
= [(1/3) * 6^3 - (5/2) * 6^2 + 6 * 6] - [(1/3) * (-3)^3 - (5/2) * (-3)^2 + 6 * (-3)]
Evaluating this expression will give you the displacement of the particle during the time interval [-3,6].
s(t) = 1/3 t^3 + 5/2 t^2 + 6t
evaluate s(6)-s(-3)
not sure what negative time involves, but hey, it's just math...