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The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8,    0 ≤ t ≤ 5(a) Find the displacement. m(b) Find the distance traveled by the particle during the given time interval. m

Question

The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8,    0 ≤ t ≤ 5(a) Find the displacement. m(b) Find the distance traveled by the particle during the given time interval. m

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Solution

(a) To find the displacement, we need to integrate the velocity function from the lower limit to the upper limit of the given interval.

The displacement (s) is the integral of the velocity function v(t) from 0 to 5.

∫ from 0 to 5 (3t - 8) dt

= [1.5t^2 - 8t] from 0 to 5

= (1.55^2 - 85) - (1.50^2 - 80)

= (37.5 - 40) - 0

= -2.5 m

So, the displacement of the particle is -2.5 meters. The negative sign indicates that the particle is 2.5 meters to the left of the starting point.

(b) To find the total distance traveled, we need to consider the absolute value of the velocity, because distance is always positive.

First, find when the particle changes direction by setting v(t) = 0.

3t - 8 = 0

t = 8/3 ≈ 2.67

So, the particle changes direction at t = 2.67 seconds.

Now, calculate the distance from 0 to 2.67 seconds and from 2.67 to 5 seconds separately by integrating the absolute value of the velocity function.

Distance = ∫ from 0 to 2.67 |3t - 8| dt + ∫ from 2.67 to 5 |3t - 8| dt

= ∫ from 0 to 2.67 (8 - 3t) dt + ∫ from 2.67 to 5 (3t - 8) dt

= [8t - 1.5t^2] from 0 to 2.67 + [1.5t^2 - 8t] from 2.67 to 5

= (21.36 - 10.7) + (37.5 - 21.35)

= 10.66 + 16.15

= 26.81 m

So, the total distance traveled by the particle is 26.81 meters.

This problem has been solved

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