Question

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6s is :

• A. $\frac{49}{4}$m
• B. 11m
• C. $\frac{37}{3}$m

Answer 1

Answer: (C)

$\frac{37}{3}$m

Answer 2

The problem asks for the total distance covered by a body in 6 seconds, given its velocity-time (v-t) graph. The total distance covered is the sum of the magnitudes of the areas under the v-t graph.

Let's break down the graph into distinct segments and calculate the area for each:

1. Segment O to A (t = 0s to t = 2s):

   • This segment represents the motion from (0,0) to (2,4).
   • The shape formed with the time axis is a triangle.
   • Base of the triangle = 2s.
   • Height of the triangle = 4 m/s.
   • Area (Distance) for this segment = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \text{s} \times 4 \text{ m/s} = 4 \text{ m}$.

2. Segment A to B (t = 2s to t = 3s):

   • This segment represents the motion from (2,4) to (3,4).
   • The shape formed with the time axis is a rectangle.
   • Base of the rectangle = 3s - 2s = 1s.
   • Height of the rectangle = 4 m/s.
   • Area (Distance) for this segment = $\text{base} \times \text{height} = 1 \text{s} \times 4 \text{ m/s} = 4 \text{ m}$.

3. Segment B to D (t = 3s to t = 6s):

   • This segment goes from (3,4) to (6,-2).

   • The problem states that point S is at 4.333 seconds. We recognize 4.333 seconds as $4 \frac{1}{3} \text{ s} = \frac{13}{3} \text{ s}$.

   • The point S is where the velocity becomes zero. So, the line segment from B to D crosses the time axis at $t = \frac{13}{3} \text{ s}$. This implies that the motion changes direction at $t = \frac{13}{3} \text{ s}$.

   • We need to calculate the area for two sub-segments: B to S (above the axis) and S to D (below the axis).

   • Sub-segment B to S (t = 3s to t = 13/3 s):

     • This forms a triangle above the time axis.
     • Base of the triangle = $\frac{13}{3} \text{ s} - 3 \text{ s} = \frac{13 - 9}{3} \text{ s} = \frac{4}{3} \text{ s}$.
     • Height of the triangle (velocity at t=3s) = 4 m/s.
     • Area (Distance) for this segment = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{4}{3} \text{ s} \times 4 \text{ m/s} = \frac{16}{6} \text{ m} = \frac{8}{3} \text{ m}$.

   • Sub-segment S to D (t = 13/3 s to t = 6s):

     • This forms a triangle below the time axis.
     • Base of the triangle = $6 \text{ s} - \frac{13}{3} \text{ s} = \frac{18 - 13}{3} \text{ s} = \frac{5}{3} \text{ s}$.
     • Magnitude of Height (velocity at t=6s) = $|-2 \text{ m/s}| = 2 \text{ m/s}$.
     • Area (Distance) for this segment = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{5}{3} \text{ s} \times 2 \text{ m/s} = \frac{5}{3} \text{ m}$.

Total Distance Covered:

Sum the areas (distances) from all segments: Total Distance = (Distance O to A) + (Distance A to B) + (Distance B to S) + (Distance S to D) Total Distance = $4 \text{ m} + 4 \text{ m} + \frac{8}{3} \text{ m} + \frac{5}{3} \text{ m}$ Total Distance = $8 \text{ m} + \frac{8+5}{3} \text{ m}$ Total Distance = $8 \text{ m} + \frac{13}{3} \text{ m}$ Total Distance = $\frac{24}{3} \text{ m} + \frac{13}{3} \text{ m}$ Total Distance = $\frac{37}{3} \text{ m}$.

The final answer is $\frac{37}{3}$m.

Explanation of the solution: The total distance covered by the body is the sum of the magnitudes of the areas under the velocity-time graph.

1. Calculate the area of the triangle from t=0s to t=2s: $\frac{1}{2} \times 2 \times 4 = 4$ m.
2. Calculate the area of the rectangle from t=2s to t=3s: $1 \times 4 = 4$ m.
3. The point S where velocity is zero is given as t = 4.333s = 13/3 s.
4. Calculate the area of the triangle from t=3s to t=13/3s (above axis): $\frac{1}{2} \times (\frac{13}{3} - 3) \times 4 = \frac{1}{2} \times \frac{4}{3} \times 4 = \frac{8}{3}$ m.
5. Calculate the area of the triangle from t=13/3s to t=6s (below axis, take magnitude): $\frac{1}{2} \times (6 - \frac{13}{3}) \times |-2| = \frac{1}{2} \times \frac{5}{3} \times 2 = \frac{5}{3}$ m.
6. Sum all the distances: $4 + 4 + \frac{8}{3} + \frac{5}{3} = 8 + \frac{13}{3} = \frac{24+13}{3} = \frac{37}{3}$ m.