- #1

sp3sp2sp

- 100

- 4

## Homework Statement

A particle starts from x_0 = 14m at t_0 = 0 and moves with the velocity graph shown

What is the object's position at t = 2 s?

## Homework Equations

Area under velocity time graph is the displacement of the particle.

## The Attempt at a Solution

The part that confuses me is where the velocity goes negative. I think that I should assign negative value to the portion where v_x is negative.

So from t=1 to t=2: (1/2)(1*4) = 2m

from t=0 to t=1 it is negative: - (1/2)(1*4) = -2m

-2m+2m = 0m = no displacement?

Is this correct before I try using calculus to solve same problem? Thanks!

sorry I forgot the initial velocity = 14m/s.

So objects position after 2s is 0m + 14m = 14m