Question

Vx (m/s) 20 10 t (s) 10 15 -10

The velocity of a container traveling along the axis is shown as a function of time in the figure below. If the container is at

x = 0

when

t = 0,

answer the following

A) What is the final position x (in m) of the object at

t = 18 s?

B) Through what total distance (in m) has the object moved between t = 0 and t = 18 s?

Answer 1

Area of velocity time graph gives net displacement. Area of OABCO = area of trapezium x1 = 1/2 (OC + AB) (h1) x_1 = 1/2 (6s + 4s) (-10 m/s) x_1 = 1/2 (10s) (-8 m/s) = -40 m Area of CDEFC = 1/2 (CF + DE) (h_2) x_2 = 1/2 (12s + 2s) (16 m/s) x_2 = 1/2 (14 sec) (16 m/s) = 112 m Total displacement = d = x_1 + x_2 = -40m + 112 m = 72 m \r\n Given at t = 0, x_0 = 0 Then consider final position = x x - x_0 = d x = d + x_0 = 62m + 0 = 62m x = 72 m (B) Total distance = Vertical-bar x_1 Vertical-bar + Vertical-bar x_2 Vertical-bar = Vertical-bar -40m Vertical-bar + Vertical-bar 112m Vertical-bar = 152m