If the following graph is given then, why is the displacement equal to the area of the shaded triangle above the axis minus the are of the shaded triangle below the $t$ axis.
Why can it not be area of the shaded triangle above the axis minus the area of the trapezium
I get that it’s because we take the area with respect to $t$ axis. But I don’t understand why.
Best Answer
Velocity is defined as $\vec v = {d \vec s \over dt}$ where $\vec s$ is position which varies with time. For one dimensional motion in say the $x$ direction $v = {dx \over dt}$. Therefore $x(t_1) - x_0 = \int_{0}^{t_1} v(t) dt$ where $x(t_1)$ is the final displacement in the $x$ direction at time $t_1$, $x_0$ is the initial position, and $v(t)$ is the time dependent velocity in the $x$ direction. $\int_{0}^{t_1} v(t) dt$ is the area under the $v(t)$ versus $t$ curve between times $0$ and $t_1$, and this area is the total $x$ displacement $x(t_1) - x_0$. For your example $v(t)$ is linear in time, initially in the positive $x$ direction which increases $x$, but later at 1 sec $v(t)$ is in the negative $x$ direction which decreases $x$; the shaded area you show is correct for the area representing the total displacement.