Question

show that area under the velocity time graph of an object moving with constant acceleration in a straight line is equal to distance covered by the object in that interval

Answer 1

To show:

Area under the velocity time graph of an object moving with constant acceleration in a straight line is equal to the distance covered by that object in that interval.

Proof:

Let us assume that an object started from rest and is moving with a constant acceleration of magnitude "a".

So, its Velocity time graph will be something like this:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(1,1){\vector(1,0){4}}\put(1,1){\vector(0,1){4}}\put(1,1){\line(1,1){3}}\put(3,0.5){t}\put(0.5,3){v}\end{picture}}$

Now , displacement of the object as per equations of kinematics will be given as:

$\therefore \: s= ut + \dfrac{1}{2} a {t}^{2}$

$= > \: s= (0)t + \dfrac{1}{2} a {t}^{2}$

$= > \: s= \dfrac{1}{2} a {t}^{2} \: \: \: \: \: .........(1)$

Now , area of under the graph:

$\therefore \: area = \dfrac{1}{2} \times base \times height$

$= > \: area = \dfrac{1}{2} \times t\times v$

$= > \: area = \dfrac{1}{2} \times t\times (u + at)$

$= > \: area = \dfrac{1}{2} \times t\times (0 + at)$

$= > \: area = \dfrac{1}{2} \times t\times at$

$= > \: area = \dfrac{1}{2} a {t}^{2} \: \: \: \: \: \: \: ........(2)$

Since (1) = (2) , we can say that:

Area under velocity-time graph gives displacement.