A particle moves along the x-axis with velocity v(t) = t^2 − 1, with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = 2 seconds.
clearly that would be
∫[0,2] t^2-1 dt = 2/3
But that is the algebraic distance (final displacement), since it starts out with negative velocity. To get the actual distance traveled, you need to separate the two parts:
∫[0,1] -(t^2-1) dt + ∫[1,2] t^2-1 dt = 2/3 + 4/3 = 2
To find the total distance traveled by the particle, we need to find the displacement at each time interval and then sum up the absolute values of the displacements.
First, let's integrate the velocity function v(t)=t^2-1 to get the position function x(t):
∫ v(t) dt = ∫ (t^2 − 1) dt
x(t) = ∫ t^2 dt - ∫ dt
x(t) = (1/3)*t^3 - t + C
Now we can find the position of the particle at t=0 and t=2 by substituting the values in the x(t) equation:
x(0) = (1/3)*0^3 - 0 + C = C
x(2) = (1/3)*2^3 - 2 + C = (8/3) - 2 + C = (2/3) + C
The displacement of the particle from t=0 to t=2 is given by the difference in positions:
Displacement = x(2) - x(0) = (2/3) + C - C = 2/3
The total distance traveled is the sum of the absolute values of the displacements. Since the displacement of 2/3 is positive, the total distance traveled is:
Total Distance = |2/3| = 2/3 feet
Therefore, the total distance traveled by the particle from t=0 to t=2 seconds is 2/3 feet.
To find the total distance traveled by the particle, we need to look at its velocity function and determine the intervals during which it is moving in the positive and negative direction.
Given the particle's velocity function v(t) = t^2 − 1, we can calculate its displacement by finding the area under the velocity-time graph.
To determine the intervals during which the particle is moving in the positive or negative direction, we need to find the points where the velocity function crosses the x-axis (where v(t) = 0). Setting v(t) = 0:
t^2 − 1 = 0
t^2 = 1
t = ±√1
t = ±1
So, the particle changes direction at t = -1 and t = 1 seconds.
Let's break down the time interval [0, 2] seconds into three segments, [0, 1], [1, 2], and [-1, 0].
From t = 0 to t = 1, the particle is moving in the positive direction, so we integrate the velocity function from 0 to 1:
∫(0 to 1) (t^2 - 1) dt
= [∫t^2 dt - ∫1 dt] (0 to 1)
= [(t^3 / 3) - t] (0 to 1)
= [(1/3 - 1) - (0 - 0)]
= [(-2/3) - 0]
= -2/3 feet
During this interval, the particle travels a distance of -2/3 feet.
From t = 1 to t = 2, the particle is moving in the negative direction, so we integrate the velocity function from 1 to 2:
∫(1 to 2) (t^2 - 1) dt
= [∫t^2 dt - ∫1 dt] (1 to 2)
= [(t^3 / 3) - t] (1 to 2)
= [(8/3 - 2) - (1/3 - 1)]
= [8/3 - 2 - 1/3 + 1]
= [11/3 - 5/3]
= 6/3
= 2 feet
During this interval, the particle travels a distance of 2 feet.
From t = -1 to t = 0, the particle is moving in the positive direction, so we integrate the velocity function from -1 to 0:
∫(-1 to 0) (t^2 - 1) dt
= [∫t^2 dt - ∫1 dt] (-1 to 0)
= [(t^3 / 3) - t] (-1 to 0)
= [(0 - 0) - (-1/3 + 1)]
= [1/3 - 1]
= [1/3 - 3/3]
= -2/3 feet
During this interval, the particle travels a distance of -2/3 feet.
Now, to find the total distance traveled, we sum up the absolute values of the distances traveled during each interval:
Total distance = | -2/3 | + | 2 | + | -2/3 |
= 2/3 + 2 + 2/3
= 2 + 2/3 + 2/3
= 4 2/3 feet
The total distance traveled by the particle from t = 0 to t = 2 seconds is 4 2/3 feet.