The velocity function is v(t)=t^2-4t+3 for a particle moving along a line. Find the displacement of the particle during the time interval [-1,6]
since displacement = velocity * time,
s = ∫[-1,6] (t^2-4t+3) dt = 70/3
To find the displacement of the particle during the time interval [-1, 6], we need to calculate the area under the velocity function curve within that interval.
We can start by finding the antiderivative (integral) of the velocity function v(t).
∫ v(t) dt = ∫ (t^2 - 4t + 3) dt
To find the antiderivative, we use the power rule of integration:
∫ t^n dt = (t^(n+1))/(n+1) + C
Applying this rule to the terms in the velocity function, we have:
∫ t^2 dt - ∫ 4t dt + ∫ 3 dt
= (t^3)/3 - 2t^2 + 3t + C
Now, we can find the displacement by evaluating this antiderivative at the endpoints of the time interval [-1, 6].
Displacement = [(t^3)/3 - 2t^2 + 3t]│[-1, 6]
= ((6^3)/3 - 2(6^2) + 3(6)) - ((-1^3)/3 - 2(-1^2) + 3(-1))
Simplifying the expression further:
= (216/3 - 2(36) + 18) - ((-1)/3 - 2 + 3)
= (72 - 72 + 18) - ((-1)/3 - 2 + 3)
= 18 - (-1/3 + 1)
= 18 - (2/3)
= (54/3) - (2/3)
= 52/3
Therefore, the displacement of the particle during the time interval [-1, 6] is 52/3 units.
To find the displacement of the particle during the given time interval [-1,6], we need to integrate the velocity function over the interval.
The displacement is the integral of the velocity function over the given time interval. So, we need to evaluate the definite integral of the velocity function v(t) between the limits -1 and 6.
Let's go step by step to solve this problem:
Step 1: Find the antiderivative of the velocity function.
To integrate the function v(t), we need to find its antiderivative. The antiderivative of t^2 is (1/3)t^3, the antiderivative of -4t is -2t^2, and the antiderivative of 3 is 3t.
So, the antiderivative of v(t)=t^2-4t+3 is:
F(t) = (1/3)t^3 - 2t^2 + 3t + C, where C is the constant of integration.
Step 2: Evaluate the definite integral using the limits.
To find the displacement, we need to evaluate the definite integral of F(t) over the interval [-1,6]. This will give us the net change in position during this time interval.
The displacement, Δd, is given by:
Δd = F(6) - F(-1)
Substituting the values into the antiderivative F(t) and calculating the displacement:
Δd = [(1/3)(6)^3 - 2(6)^2 + 3(6)] - [(1/3)(-1)^3 - 2(-1)^2 + 3(-1)]
= [(1/3)(216) - 2(36) + 18] - [(1/3)(-1) - 2(1) - 3]
= [72 - 72 + 18] - [-1/3 - 2 - 3]
= 18 + 6/3 + 3
= 18 + 2 + 3
= 23 + 3
= 26 units
Therefore, the displacement of the particle during the time interval [-1,6] is 26 units.