Question

The area under velocity-time graph gives:
A. Acceleration
B. Distance
C. Displacement
D. Velocity

Answer 1

Hint: We will see an arbitrary Velocity-time graph and then will try to find the area under this graph. Then, we have to put the expression of velocity there so that we can find the answer. The answer has to be correct by analysis of units as well.

Formula used: $A=\int\limits_{x_1}^{x_2}y.dx $

Complete step-by-step solution:
The area under a curve in x-y graph from $x_1$ to $x_2$ is given by,
$A=\int\limits_{x_1}^{x_2}y.dx $

Here, y is usually a function of x. Hence, the area under a curve can be found from any point to another point. Similarly, in case of the velocity-time graph, the area under the curve is given as, $\vec A_{vt}=\int\limits_{t_1}^{t_2}\vec v.dt $
Here, the area $\vec A_{vt}$ is given in vector form because the velocity is a vector quantity.
Now, we know one thing : $\vec v=\dfrac{d\vec r}{dt}$ . Here, $\vec r$ is the displacement from a particular point (say the origin). So, $\vec r$ is not related to how a body reached a position, but it is just the linear distance of the position from the starting point along with a direction. Now, putting the expression of $\vec v$ in the above equation, we get,
$\vec A_{vt}=\int\limits_{t_1}^{t_2}\vec v.dt=\int\limits_{t_1}^{t_2}\dfrac{d\vec r}{dt}.dt \\ =\int\limits_{\vec r_1}^{\vec r_2}d\vec r=\vec r_2-\vec r_1 $
Hence, the required area is basically $\vec r_2-\vec r_1$. In the language of science, it is called displacement i.e. the “ultimate distance” that has been travelled in the given time. So, option C is the correct answer.

Additional information:
The slope of the velocity-time graph is $\dfrac{d\vec v}{dt}$ and it is called acceleration. In a similar way the slope of the displacement-time graph i.e. $\dfrac{\vec r}{dt}$ is called the velocity.

Note: Remember the following points:
1. Distance is the length of ground that has been travelled. But, displacement is the ultimate separation between two positions. Distance is scalar but displacement is vector.
2. The area $A_{vt}$ is not an area in true sense. It’s just the area under a particular curve.