Question

Why is the area under the Velocity time graph the distance?

Answer 1

Hint: The displacement of moving objects with constant velocity is equal to the product of the object velocity and the amount of time the object is in motion.
We need a velocity-time graph when the object’s velocity gets changed.
The area under the velocity-time graph is known as displacement.

Complete step-by-step solution:
Let us draw the distance-time graph of the body from the position $$x_1$$ and $$x_2$$ as drawn below,

In the above diagram, the area under the curve is equal to the area of the triangle $$BCE$$ and the area of the rectangle ABCD, this can be expressed as follows,
$A=(AB\times AD)+({\displaystyle \frac{1}{2}}\times BC\times CE)$
Then this equation becomes,
$A=(x_1)(t_2-t_1)+\left({\displaystyle \frac{1}{2}}(t_2-t_1)(x_2-x_1)\right)$
We have to solve the above equation then it becomes,
$A=(t_2-t_1)\left(x_1+{\displaystyle \frac{1}{2}}{x}_2-{\displaystyle \frac{1}{2}}{x}_1\right)$
After simplification the equations are,
$A=(t_2-t_1)\left({\displaystyle \frac{x_1+x_2}{2}}\right)$
In the above equations When we consider the unit of term on the right-hand side, it gives
$A=meter\times \sec$
From this area under the distance-time graph gives nothing, Now draw the graph of the velocity of the body concerning the time

The expression for the area under the curve is as follows,
$A=(AB\times AD)+({\displaystyle \frac{1}{2}}\times BC\times CE)$
The equation becomes,
$A=(v_1)(t_2-t_1)+\left({\displaystyle \frac{1}{2}}(t_2-t_1)(v_2-v_1)\right)$
after solving the above equation,
$$A=(t_2-t_1)\left(v_1+{\displaystyle \frac{1}{2}}{v}_2-{\displaystyle \frac{1}{2}}{v}_1\right)$$
Hence it becomes,
$$A=(t_2-t_1)\left({\displaystyle \frac{v_1+v_2}{2}}\right)$$
In the above equation, the right-hand side determines the unit we get,
$$A={\displaystyle \frac{meter}{\sec }}\times \sec$$
$$A=meter$$
Here, the area under the curve of the velocity-time graph gives the distance covered by the object

Note:The velocity of the body is determined by the gradient curve in the distance-time graph.
The acceleration of the body is determined by the gradient curve in the velocity-time graph.
By integrating the curve we can calculate the area under the curve.