
Why is the area under the Velocity time graph the distance?
Answer
429.3k+ views
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,
$A = (AB \times AD) + (\dfrac{1}{2} \times BC \times CE)$
Then this equation becomes,
$A = ({x_1})({t_2} - {t_1}) + \left( {\dfrac{1}{2}({t_2} - {t_1})({x_2} - {x_1})} \right)$
We have to solve the above equation then it becomes,
$A = ({t_2} - {t_1})\left( {{x_1} + \dfrac{1}{2}{x_2} - \dfrac{1}{2}{x_1}} \right)$
After simplification the equations are,
$A = ({t_2} - {t_1})\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)$
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,
$A = (AB \times AD) + (\dfrac{1}{2} \times BC \times CE)$
The equation becomes,
$A = ({v_1})({t_2} - {t_1}) + \left( {\dfrac{1}{2}({t_2} - {t_1})({v_2} - {v_1})} \right)$
after solving the above equation,
\[A = ({t_2} - {t_1})\left( {{v_1} + \dfrac{1}{2}{v_2} - \dfrac{1}{2}{v_1}} \right)\]
Hence it becomes,
\[A = ({t_2} - {t_1})\left( {\dfrac{{{v_1} + {v_2}}}{2}} \right)\]
In the above equation, the right-hand side determines the unit we get,
\[A = \dfrac{{meter}}{{\sec }} \times \sec \]
\[A = meter\]
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.
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,
$A = (AB \times AD) + (\dfrac{1}{2} \times BC \times CE)$
Then this equation becomes,
$A = ({x_1})({t_2} - {t_1}) + \left( {\dfrac{1}{2}({t_2} - {t_1})({x_2} - {x_1})} \right)$
We have to solve the above equation then it becomes,
$A = ({t_2} - {t_1})\left( {{x_1} + \dfrac{1}{2}{x_2} - \dfrac{1}{2}{x_1}} \right)$
After simplification the equations are,
$A = ({t_2} - {t_1})\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)$
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,
$A = (AB \times AD) + (\dfrac{1}{2} \times BC \times CE)$
The equation becomes,
$A = ({v_1})({t_2} - {t_1}) + \left( {\dfrac{1}{2}({t_2} - {t_1})({v_2} - {v_1})} \right)$
after solving the above equation,
\[A = ({t_2} - {t_1})\left( {{v_1} + \dfrac{1}{2}{v_2} - \dfrac{1}{2}{v_1}} \right)\]
Hence it becomes,
\[A = ({t_2} - {t_1})\left( {\dfrac{{{v_1} + {v_2}}}{2}} \right)\]
In the above equation, the right-hand side determines the unit we get,
\[A = \dfrac{{meter}}{{\sec }} \times \sec \]
\[A = meter\]
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.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

The non protein part of an enzyme is a A Prosthetic class 11 biology CBSE

Which of the following blood vessels in the circulatory class 11 biology CBSE

What is a zygomorphic flower Give example class 11 biology CBSE

The deoxygenated blood from the hind limbs of the frog class 11 biology CBSE

What is the function of copulatory pads in the forelimbs class 11 biology CBSE
