The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t^2 − 2t − 15, 1 ≤ t ≤ 7
(a) Find the displacement.
(b) Find the distance traveled by the particle during the given time interval.
Check you other posts, the questions are related.
Derivative of displacement (wrt time) is velocity.
Thus, integral of velocity is displacement.
D(t) = ∫(t^2 − 2t − 15) dt
D(t) = ?
After you solved for the integration, evaluate this from time t=1 to t=7.
To find the displacement and distance traveled by the particle, we must first understand the relationship between velocity, displacement, and distance.
Velocity: The velocity of an object is the rate at which it changes its position with respect to time. It is the derivative of the position function.
Displacement: The displacement of an object is the change in its position from the initial position to the final position. It is the integral of the velocity function.
Distance: The distance traveled by an object is the total length of its path, regardless of the direction. It is the integral of the absolute value of the velocity function.
Now, let's solve each part of the question step by step.
(a) Finding the Displacement:
To find the displacement, we need to integrate the velocity function over the given time interval.
∫(t^2 - 2t - 15) dt, from t = 1 to t = 7
To integrate the velocity function, we apply the power rule of integration:
= (1/3)t^3 - t^2 - 15t + C
Next, we evaluate the definite integral by substituting the upper and lower limits of integration:
= [(1/3)(7^3) - (7^2) - 15(7)] - [(1/3)(1^3) - (1^2) - 15(1)]
Simplifying the expression, we find the displacement of the particle:
= (343/3 - 49 - 105) - (1/3 - 1 - 15) = -343/3 + 49 + 105 + 1/3 - 1 + 15
= -343/3 + 49 + 105 + 1/3 - 1 + 15 = 31
Therefore, the displacement of the particle is 31 meters.
(b) Finding the Distance Traveled:
To find the distance traveled, we need to integrate the absolute value of the velocity function over the given time interval.
∫|t^2 - 2t - 15| dt, from t = 1 to t = 7
We first identify the intervals where the velocity function is positive and negative. Looking at the given velocity function, we can see that it changes sign at t = -3 and t = 5.
Our integral can be divided into three parts:
1. ∫[(t^2 - 2t - 15)] dt, from t = 1 to t = -3
2. ∫[-(t^2 - 2t - 15)] dt, from t = -3 to t = 5
3. ∫[(t^2 - 2t - 15)] dt, from t = 5 to t = 7
For the first and third parts, we integrate as we did in part (a). For the second part, we integrate the negation of the velocity function:
= (1/3)(-3)^3 - (-3)^2 - 15(-3) - [(1/3)(1^3) - (1^2) - 15(1)] + (1/3)(7^3) - (7^2) - 15(7)
Simplifying the expression, we find the distance traveled by the particle:
= (5/3) - 3 - 15(3) - (1/3 - 1 - 15) + (343/3 - 49 - 105)
= 5/3 - 3 - 45 + 1/3 -1 + 15 + 343/3 - 49 - 105
= -10/3 + 5/3 - 1 + 15 + 343/3 - 49 - 105
= 333/3 - 145 - 1 = 111 - 145 - 1
= -35 - 1 = -36
Therefore, the distance traveled by the particle during the given time interval is 36 meters.