Animated Solution for Physics - Kinematics: The speed versus time graph for a particle is shown in the figure. The distance travelled (in metre) by the particle during the time interval $t=0\text{ s}$ to $t=5\text{ s}$ will be ……… .

When you encounter a kinematics problem involving graphs, consider it a gift! Graphical questions in JEE are often highly scoring if you know how to read the visual data. In this problem, we are given a speed-time graph for a moving particle and asked to find the total distance travelled between $t=0$ and $t=5\text{ s}$.

The graph shows a straight line starting from the origin $(0,0)$ and ending at a point corresponding to $t=5\text{ s}$. By carefully observing the y-axis, we can see that at $t=5\text{ s}$, the speed of the particle is $8{\text{ ms}}^{-1}$.

Area Under the Curve

The fundamental principle you must always remember is this: The area under a speed-time graph gives the total distance travelled by the body.

Why is this true? Because distance is the integral of speed with respect to time ($S=\int vdt$), and geometrically, the definite integral represents the area under the curve.

Let's focus on the region bounded by the graph, the x-axis, and the vertical line at $t=5\text{ s}$. This region forms a perfect right-angled triangle, which we can name $\Delta OAB$, where: - $O$ is the origin $(0,0)$ - $B$ is the point on the time axis $(5,0)$ - $A$ is the peak point on the graph $(5,8)$

To find the area of this triangle, we use the standard geometry formula:

From our coordinates, the dimensions are straightforward: - Base ($OB$): The time interval, which is $5\text{ s}$. - Height ($AB$): The maximum speed reached, which is $8{\text{ ms}}^{-1}$.

Substituting these values into our formula:

And there we have it! The particle travelled a total distance of $20\text{ m}$.

Always pay close attention to the axis labels. This was a speed-time graph, so the area directly gave us the distance. If it had been a velocity-time graph and a portion of the graph dipped below the x-axis, that area would be considered negative when calculating displacement, but you would still take its absolute (positive) value when calculating the total distance. Keep your concepts crystal clear, and you'll never fall for these traps!