A particle moves in a straight line with velocity t^(-2)-(1/16) ft/s. Find the total displacement and total distance traveled over the time interval [2,5
ok i got the displacement is .1125 ft
however i can not find the distance! help!
For the displacement I got the same answer.
For the total distance traveled, you have to find the length of the curve.
This topic is not usually taught in highschool Calculus, what country are you in?
Here is a link you can use as an example for your problem.
http://www.mathwords.com/a/arc_length_of_a_curve.htm
To find the total distance traveled over the time interval [2, 5], we need to calculate the distance traveled during each sub-interval where the velocity is positive and determine the distance traveled during each sub-interval where the velocity is negative.
Let's start by finding the sub-intervals where the velocity is positive:
For t = 2, the velocity is given by:
v(2) = (2)^(-2) - (1/16) = 1/4 - 1/16 = 3/16 ft/s
For t = 5, the velocity is given by:
v(5) = (5)^(-2) - (1/16) = 1/25 - 1/16 = -9/400 ft/s
We can observe that the velocity is positive between t = 2 and t = 5.
Now let's find the sub-intervals where the velocity is negative:
For t = 2, the velocity is negative:
v(2) = 3/16 ft/s
For t = 4, the velocity is given by:
v(4) = (4)^(-2) - (1/16) = 1/16 - 1/16 = 0 ft/s
We can observe that the velocity is negative between t = 2 and t = 4.
Now let's calculate the distance traveled during each sub-interval where the velocity is positive:
Distance traveled from t = 2 to t = 5:
d1 = ∫[2,5] v(t) dt
= ∫[2,5] (t^(-2) - 1/16) dt
= [-t^(-1) - 1/16t] from 2 to 5
= [-(1/5) - 1/80] - [-(1/2) - 1/32]
= -(1/5) - 1/80 + 1/2 + 1/32
= -17/80 + 21/32
= 33/160 - 105/160
= -72/160
= -9/20 ft
Since the distance cannot be negative, we take the absolute value:
|d1| = 9/20 ft
And calculate the distance traveled during the sub-interval where the velocity is negative:
Distance traveled from t = 2 to t = 4:
d2 = ∫[2,4] v(t) dt
= ∫[2,4] (3/16) dt
= (3/16) * [t] from 2 to 4
= (3/16) * (4 - 2)
= 3/8 ft
Finally, we find the total distance traveled by summing the absolute values of the distances:
Total distance traveled = |d1| + d2
= 9/20 + 3/8
= (9/20)*(4/4) + (3/8)*(5/5)
= 36/80 + 15/40
= 51/80 ft
Therefore, the total distance traveled over the time interval [2, 5] is 51/80 ft.
To find the total distance traveled by the particle over the time interval [2,5], you need to integrate the absolute value of the velocity function over that interval.
Given that the velocity function is v(t) = t^(-2) - 1/16 ft/s, we can start by finding the absolute value of this function.
|v(t)| = |t^(-2) - 1/16|
Next, we integrate this absolute value over the time interval [2,5].
Total distance = ∫[2,5] |v(t)| dt
To evaluate this integral, we need to split it into two parts based on the discontinuity at t = 0. The integral from 2 to 5 will be the sum of two integrals:
Total distance = ∫[2,5] t^(-2) dt + ∫[2,5] 1/16 dt
Integrating each part separately:
∫[2,5] t^(-2) dt = [-t^(-1)]|[2,5] = (-1/5) - (-1/2) = 3/10
∫[2,5] 1/16 dt = (1/16) * (t)|[2,5] = (5/16) - (2/16) = 3/16
Putting the parts back together:
Total distance = (3/10) + (3/16) = 48/80 + 15/80 = 63/80 ft
Therefore, the total distance traveled by the particle over the time interval [2,5] is 63/80 ft.