# The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the displacement. ___________m (b) Find the distance traveled by the particle during the given time interval. ___________m

You need to evaluate the definite integral of velocity function over interval [0,3] such that:

`int_0^3 (5t - 9)dt = int_0^3 5t dt - int_0^3 9dt`

`int_0^3 (5t - 9)dt = (5t^2/2 - 9t)|_0^3`

`int_0^3 (5t - 9)dt = 45/2 - 27`

`int_0^3 (5t - 9)dt = (45 - 54)/2`

`int_0^3 (5t - 9)dt = -9/2`

`int_0^3 (5t - 9)dt = -4.5`

Hence, evaluating displacement of particle moving along the line yields `Delta x = -4.5 m.`

b) You need to determine the interval the velocity is negative and the interval velocity is positive, such that:

`5t-9 = 0 =gt 5t=9 =gt t=9/5`

You need to notice that the values of velocity function are negative over `[0,9/5]`  and positive over `[9/5,3], ` hence you need to evaluate the absolute values of the definite integrals:

`|int_0^(9/5) (5t - 9) dt| + |int_(9/5)^3 (5t - 9) dt| `

`|(5t^2/2 - 9t)|_0^(9/5)| + |(5t^2/2 - 9t)|_(9/5)^3|`

`|81/10 - 81/5| + |45/2 - 81/10 - 27 + 81/5|`

`81/10 + |81/10 + 45/2 - 27|`

`81/10 + 36/10 = 117/10`

Hence, evaluating the distance travelled by particle ovr interval [0,3] yields d = 11.7 m.

Approved by eNotes Editorial Team