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.