Question

The velocity function (in meters per second) for a particle moving along a line is v(t)=5t−8,0≤t≤3. (5.4) a. Find the displacement ∫03​v(t)dt. b. Find the distance traveled ∫03​∣v(t)∣dt

Answer 1

The displacement of the particle is -7.5 meters and the distance traveled is 13.9 meters.

Step-by-step explanation:

To find the displacement of the particle, we need to evaluate the integral of the velocity function over the given time interval of 0≤t≤3:

∫03v(t)dt = ∫03(5t-8)dt

Using the power rule of integration, we can find the antiderivative of 5t - 8:

∫03(5t-8)dt = [2.5t2 - 8t]03 = (2.5(3)2 - 8(3)) - (2.5(0)2 - 8(0)) = 16.5 - 24 = -7.5 meters

Therefore, the displacement of the particle is -7.5 meters.

To find the distance traveled, we need to evaluate the integral of the absolute value of the velocity function over the same time interval:

∫03|v(t)|dt = ∫03|5t-8|dt

Since the absolute value of a function is non-negative, we can split the integral at the point where the function changes sign:

∫03|v(t)|dt = ∫08/5(8-5t)dt + ∫8/53(5t-8)dt

Using the power rule of integration, we can evaluate each integral separately:

∫8/53(5t-8)dt = [2.5t2 - 8t]8/53 = (2.5(3)2 - 8(3)) - (2.5(8/5)2 - 8(8/5)) = 16.5 - 17.6 = -1.1 meters

Therefore, the distance traveled by the particle is |12.8| + |-1.1| = 13.9 meters.

Answer 2

a. The displacement is -1.5 meters. b. The distance traveled is 1.5 meters.

Step-by-step explanation:

a. To find the displacement, we need to integrate the velocity function over the given interval. The integral of 5t - 8 with respect to t from 0 to 3 is [ (5/2)
`t^2` - 8t ] evaluated from 0 to 3. Plugging in the upper and lower limits, we get [ (5/2)
`(3)^2` - 8(3) ] - [ (5/2)
`(0)^2`- 8(0) ] = [ (5/2)(9) - 24 ] - 0 = 22.5 - 24 = -1.5 meters.

b. To find the distance traveled, we need to calculate the integral of the absolute value of the velocity function over the given interval. The integral of |5t - 8| with respect to t from 0 to 3 is [ (5t^2/2 - 8t) ] evaluated from 0 to 3. Plugging in the upper and lower limits and taking the absolute value, we get | (5(3)^2/2 - 8(3)) | - | (5(
`0)^{2/2` - 8(0)) | = | (45/2 - 24) | - 0 = | (45/2 - 24) | = | (45/2 - 48/2) | = | (-3/2) | = 3/2 = 1.5 meters.