Distances & Areas, Left/Right/Midpoint Sums

Question

The graph below shows the velocity of an object (in ft/sec) as a function of time 𝑡 (in seconds).

(Explanation of the graph)

The y axis goes from -4 to 4 The x axis goes from -1 to 7

There are points at (0,3), (3,2), (5,0), (6,-3), and (7,-3). Straight lines connect all the points.

Find the total displacement from 𝑡=0 to 𝑡=3:

Displacement =  ?  feet

Find the total displacement from 𝑡=3 to 𝑡=7:

Displacement =  ? feet

Answer 1

(i). Displacement from x=0 to x=3 by Mid point Method:

Equation of line joinig points A(0,3)and B(3,2): with intervals (0,1), (1,2) and (2,3). δx = 1

y - 3 = [(2-3)/(3-0)](x- 0)

y-3 = -x/3, y(0.5) = 3 - (0.5/3) = 2.833, y(1.5) = 3 - 1.5/3 = 2.5, y(2.5/3) = 3 - 2.5/3 = 2.167

Sm = Sum by mid point method: = [y(0.5)* + y(1.5) + y(2.5)] δx = 2.833 + 2.50 + 2.167 = 7.5

(ii). Displacement from x=3 to x=7 by Mid point Method: Consider intervals (3,5), (5,6), ^,70

Equation of line BC joining points B(3,2) and C(5, 0):

y - 2 = [(0-2)/(5-3)] (x - 3)

y -2 = -2/3(x - 3) ⇒ y = -(2/3)x + 4,

y(4) = -(2/3)(4) +4 = 1.333, value of y at mid point of interval (3,5)

Equation of line CD joining the points (5,0) and (6, -3)

y - 0 = [(-3 - 0)/(6 - 5)](x-5) ⇒ y = -3(x - 5) ⇒ y = -3x + 15

y (5.5) = value of ordinate y at mid point of interal (5, 6) = -3(5.5) +15 = -16.5 +15 = -1.5

Equation of line DE: y = -3

y(6.5) = -3

Displacement ( from x=3 to x =7) =y(4)* δx1 + y(5.5)* δx2 + y(6.5)*δx3

= 1.333*2 - (-1.5)*1 - (-3)*1

= 2.666 + 1.5 +3 = 7.17

Find the total displacement from 𝑡=0 to 𝑡=3:

Displacement =  7.5  feet

Find the total displacement from 𝑡=3 to 𝑡=7:

Displacement =  7.17 feet

Answer 2