A particle moves according to the equation of motion s(t)=t^2-2t+3 where s(t) is measured in feet and t is measured in seconds. Find the total distance traveled in the 3 seconds.
If you just integrate the velocity, you wind up with the final displacement, not the distance traveled. So, to avoid subtracting backward motion, use the absolute value of the velocity:
v(t) = 2t-2
distance = ∫[0,3] |2t-2| dt
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,3%5D+%7C2t-2%7C+dt
Well, well, well, let's have some mathematical fun, shall we?
To find the total distance traveled, we need to look at the absolute value of the velocity function. The velocity function is the derivative of the position function, which in this case is s(t).
So, let's take the derivative of s(t) to get v(t).
s(t) = t^2 - 2t + 3
Taking the derivative:
v(t) = 2t - 2
Now, to find the total distance traveled in 3 seconds, we need to integrate the absolute value of v(t) from t = 0 to t = 3.
∫[0 to 3] |v(t)| dt
Let's break it down into two parts. First, let's integrate from t = 0 to t = 1:
∫[0 to 1] |2t - 2| dt
As you know, integrating absolute values can be quite tricky. We need to consider the cases where (2t - 2) is positive and negative separately.
For (2t - 2) ≥ 0, solve 2t - 2 = 0, we get t = 1.
So, for t ∈ [0, 1], |2t - 2| = 2t - 2.
Now, let's integrate:
∫[0 to 1] (2t - 2) dt = t^2 - 2t |[0 to 1] = (1^2 - 2(1)) - (0^2 - 2(0)) = -1
Now, let's integrate from t = 1 to t = 3:
∫[1 to 3] |2t - 2| dt
For (2t - 2) < 0, solve 2t - 2 = 0, we get t = 1.
So, for t ∈ [1, 3], |2t - 2| = -(2t - 2) = -2t + 2.
Integrating again:
∫[1 to 3] (-2t + 2) dt = -t^2 + 2t |[1 to 3] = (-(3^2) + 2(3)) - (-(1^2) + 2(1)) = -4
Now, to find the total distance, we add the absolute values of both calculations:
|-1| + |-4| = 1 + 4 = 5
And there you have it! The total distance traveled in 3 seconds is 5 feet.
To find the total distance traveled in 3 seconds, we need to calculate the distance traveled between each interval where the particle changes direction.
First, we find the time when the particle changes direction by setting the velocity equation equal to zero:
v(t) = s'(t) = 2t - 2 = 0
Solving for t, we get:
2t - 2 = 0
2t = 2
t = 1
So at t = 1 second, the particle changes direction.
Now, let's calculate the distance traveled in each interval.
Interval 1: t = 0 to t = 1 second
s(t) = t^2 - 2t + 3
To find the distance traveled, we calculate the difference in position at t = 1 and t = 0:
d1 = |s(1) - s(0)|
= |(1^2 - 2(1) + 3) - (0^2 - 2(0) + 3)|
= |(1 - 2 + 3) - (0 - 0 + 3)|
= |2 - 3|
= 1
Interval 2: t = 1 to t = 3 seconds
s(t) = t^2 - 2t + 3
To find the distance traveled, we calculate the difference in position at t = 3 and t = 1:
d2 = |s(3) - s(1)|
= |(3^2 - 2(3) + 3) - (1^2 - 2(1) + 3)|
= |(9 - 6 + 3) - (1 - 2 + 3)|
= |6 - 2|
= 4
Now, we can find the total distance traveled:
Total distance = d1 + d2
= 1 + 4
= 5 feet
Therefore, the total distance traveled in the 3 seconds is 5 feet.
To find the total distance traveled, we need to consider both the positive and negative distances covered by the particle.
First, let's find the distance covered during the first 3 seconds. We can calculate the distance traveled by subtracting the initial position from the final position. So, the distance covered in the first 3 seconds (d1) is given by:
d1 = s(3) - s(0)
To calculate s(3), substitute t = 3 into the equation of motion:
s(3) = (3^2) - 2(3) + 3
= 9 - 6 + 3
= 6
And s(0) is the initial position, which is equal to s(0) = (0^2) - 2(0) + 3 = 3.
Therefore,
d1 = s(3) - s(0)
= 6 - 3
= 3 feet
So, the particle traveled 3 feet in the first 3 seconds.
To find the total distance traveled, we also need to consider the distance covered in the time interval when the particle moved in the opposite direction (in this case, when the particle moved from t = 3 to t = 0). In other words, we need to find the absolute value of the distance covered during this interval.
The distance covered from t = 3 to t = 0 (d2) is given by:
d2 = |s(0) - s(3)|
Using the previously calculated values, we have:
d2 = |3 - 6|
= |-3|
= 3 feet
Since the particle traveled 3 feet in the opposite direction from t = 3 to t = 0, we take the absolute value to find the distance covered.
To find the total distance traveled, we sum up the distances for the two intervals:
Total distance traveled = d1 + d2
= 3 + 3
= 6 feet
Therefore, the total distance traveled by the particle in the first 3 seconds is 6 feet.