Question

What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer 1

Hint: In order to solve this question we need to understand distance, displacement and velocity. Distance is defined as the total path length travelled between initial and final position where displacement is defined as length of shortest path travelled between initial and final points. Speed is defined as the distance covered per unit time, it is a scalar quantity. Velocity is defined as the displacement of a body in unit interval of time, it is vector quantity.

Complete step by step answer:
Since we know the three equation of motion given by,
$v = u + \dfrac{1}{2}at \to (i)$
And, $s = ut + \dfrac{1}{2}a{t^2} \to (ii)$
And, ${v^2} = {u^2} + 2as \to (iii)$
Here “v” is final speed of body, “u” is initial speed of body, “a” is acceleration of body and “t” is time

We know we can use these equations only when the acceleration of a body is uniform in time.But if this is not condition then, we use differential forms of velocity and acceleration to calculate equation of motion $v = \dfrac{{dx}}{{dt}} \to (iv)$ and $a = \dfrac{{dv}}{{dt}} \to (v)$

Physically it means that velocity is slope of displacement time graph and acceleration is slope of velocity time graph.Also we can use this equation to find the displacement and velocity as,
From (iv) we get, $x = \int {vdt} $ and
From (v) we get, $v = \int {(a)dt} $
Here “x” is displacement, “v” is velocity and “a” is acceleration

Since integration physically represents area, so area under velocity time graph is displacement.

Note:It should be remembered that the area calculated is approximated displacement and it must be found under certain limits because to calculate are we have assumed an infinitesimally time $dt$ multiplied with velocity to calculate displacement $dx$ and then later integrate it to find whole displacement.