The acceleration of a particle moving in a straight line is given by a(t) = 6(t-3). (b) Find the particle’s displacement during the time interval 0 less than or equal to t is less than or equal to 5.
You need to remember the following relations between acceleration, velocity and displacement such that:
`v(t) = int a(t) dt`
`s(t) = int v(t) dt`
Hence, you need to find the velocity first and then you may evaluate the displacement of particle, over the interval [0,5], such that:
`v(t) = int_0^5 6(t - 3)dt => v(t) = 6int_0^5 (t - 3) dt`
You should split the integral in two, using the property of linearity, such that:
`v(t) = 6int_0^5 (t) dt- 6int_0^5(3) dt`
Using the fundamental theorem of calculus yields:
`v(t) = 6(t^2/2 - 3t)|_0^5 => v(t) = 3*5^2 - 18*5 - 0 + 0`
`v(t) = -15`
`s(t) = |int_0^5 v(t) dt| => s(t) = |- int_0^5 15 dt| => s(t) = 15t|_0^5`
`s(t) = 15*5- 15*0 => s(t) = 75` units
Hence, evaluating the displacement, under the given conditions, yields `s(t) = 75` units.
check Approved by eNotes Editorial