The velocity function is v(t) = t^2 - 6 t + 8 for a particle moving along a line. Find the displacement (net d

Question

The velocity function is v(t) = t^2 - 6 t + 8 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [-3,6].

displacement =

Answer 1

d(t) = ∫ t² - 6t + 8 dt
d(t) = t³/3 - 3t² + 8t between - 3 and 6
d(t) = (72 - 108 + 48) - (-9 - 27 - 24)
d(t) = 12 - ( - 60)
d(t) = 72

Answer 2

Displacement = Integral of [ t^2 - 6 t + 8 ] within limits (-3,6)

= [t^3 / 3 - 6/2 ( t^2) + 8t] ( -3, 6)
= [6^3 / 3 - 3 (36) + (8x6)] - [ (-3^3)/3 - 3(-3)^2 + (-3) ]
= 72 units ............... Answer

Answer 3

you should drive than that function wich is x(t)=2t-6
and then use -3 and 6 to find the displacemnt.