Animated Solution for Physics - Kinematics: The v-t graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 s. The total distance covered by the body in 6 s is

Unlocking Distance from Velocity-Time Graphs

When you look at a velocity-time ($v-t$) graph, you are looking at a visual story of a particle's journey. The slope tells you about acceleration, but the area under the curve holds the secret to how far the particle has traveled.

However, there is a crucial distinction you must always keep in mind: Distance vs. Displacement.

Displacement is a vector. It cares about direction. If a particle moves forward and then backward, the backward motion subtracts from the total displacement. On a $v-t$ graph, this means areas below the time axis (where velocity is negative) are treated as negative.

Distance, on the other hand, is a scalar. It is the total path length covered, regardless of direction. Therefore, to find the distance, we must take the absolute value (magnitude) of all areas, whether they lie above or below the time axis.

To calculate the total area, we break the complex polygon into simple, manageable geometric shapes: triangles and rectangles. Let's analyze the journey from $t=0$ to $t=6$ seconds by dividing it into five distinct regions.

Region 1: The First Triangle ($A_1$) From $t=0$ to $t=2$ s, the graph forms a triangle $\Delta OAE$. - Base = $2$ s - Height = $4$ m/s

Region 2: The Rectangle ($A_2$) From $t=2$ to $t=3$ s, the velocity is constant, forming a rectangle $ABFE$. - Base = $3-2=1$ s - Height = $4$ m/s

Region 3: The Decelerating Triangle ($A_3$) From $t=3$ to $t=4.333$ s, the particle slows down to rest. The point $S$ is at $4.333$ s, which is exactly $\frac{13}{3}$ s. This forms triangle $\Delta BSF$. - Base = $\frac{13}{3}-3=\frac{4}{3}$ s - Height = $4$ m/s

Handling Negative Velocity

Region 4: The Negative Triangle ($A_4$) From $t=4.333$ s to $t=5$ s, the velocity becomes negative. This forms triangle $\Delta SCG$ below the axis. Since we are calculating distance, we take the magnitude of the height. - Base = $5-\frac{13}{3}=\frac{2}{3}$ s - Height magnitude = $|-2|=2$ m/s

Region 5: The Final Triangle ($A_5$) From $t=5$ to $t=6$ s, the particle returns to rest, forming triangle $\Delta GCD$. - Base = $6-5=1$ s - Height magnitude = $|-2|=2$ m/s

Now, we simply add the magnitudes of all these areas together to find the total distance covered in the 6 seconds.

Let's group the integers and the fractions to make the addition easier:

The total distance covered by the body is $\frac{37}{3}$ meters. Always remember to read the question carefully to see if it asks for distance or displacement, as that single word completely changes how you handle the areas below the time axis!