The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 8, 0 ≤ t ≤ 3?

(a) Remember that displacement is the distance that the particle is from it's starting position

i.e. If travelling to the right is positive, then a displacement of -3/2 m is 3/2 m to the left

Our critical point is when the velocity is zero

i.e. when 5t - 8 = 0

so, when t = 1.6

We need to evaluate the distance travelled from t = 0 to t = 1.6 and separately from t = 1.6 to t = 3

Now, s(t) = 5t²/2 - 8t

From, t = 0 to t = 1.6 we get:

5(1.6)²/2 - 8(1.6) => -6.4....i.e. 6.4 m to the left

From t = 1.6 to t = 3 we get:

[5(3)²/2 - 8(3)] - [5(1.6)²/2 - 8(1.6)]

i.e. (-1.5) - (-6.4) => -1.5 + 6.4 = 4.9 m to the right

so, total distance is 6.4 + 4.9 = 11.3 m

Note: If we just go from t = 0 to t = 3 in one calculation we get:

-6.4 + 4.9 = -1.5....which is displacement in part a.

In summary we ignore -ve signs for displacement when calculating total distance.

e.g. 3 m left then 5 m right => |-3| + 5 = 3 + 5 = 8

:)> 

∫(5*t-8,t,0,3)= 5t^2/2-8t, 0,3]=  [5*3^2/2 -8*3] -[0]= 45/2 - 48/2 = -3/2m