The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the…
The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between $t=4 \mathrm{~s}$ and $\mathrm{t}=6 \mathrm{~s}$ is ____
$0.5 \mathrm{~ms}^{-1}$
$2.5 \mathrm{~ms}^{-1}$
$7.5 \mathrm{~ms}^{-1}$
$9.5 \mathrm{~ms}^{-1}$
Solution
Step 1: Calculate the displacement To find the displacement from a velocity-time graph, we need to calculate the area under the curve between the specified times. The area under the graph from \(t=4\mathrm{~s}\) to \(t=6\mathrm{~s}\) can be split into two parts: a triangle below the x-axis and a triangle above the x-axis. Area 1 (from t=4s to t=5s): This is a right-angled triangle with base \((5-4)=1\mathrm{~s}\) and height \(-2\mathrm{~m/s}\).\(\text{Area}_{1}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 1\mathrm{~s}\times (-2\mathrm{~m/s})=-1\mathrm{~m}\)The negative sign indicates displacement in the negative direction. Area 2 (from t=5s to t=6s): This is a right-angled triangle with base \((6-5)=1\mathrm{~s}\) and height \(4\mathrm{~m/s}\).\(\text{Area}_{2}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 1\mathrm{~s}\times 4\mathrm{~m/s}=2\mathrm{~m}\)The total displacement is the sum of these areas:\(\text{Displacement}=\text{Area}_{1}+\text{Area}_{2}=-1\mathrm{~m}+2\mathrm{~m}=1\mathrm{~m}\)
Step 2: Calculate the time interval The time interval is the difference between the final and initial times.\(\text{Time\ Interval}=6\mathrm{~s}-4\mathrm{~s}=2\mathrm{~s}\)
Step 3: Calculate the average velocity Average velocity is defined as the total displacement divided by the time interval.\(\text{Average\ Velocity}=\frac{\text{Total\ Displacement}}{\text{Time\ Interval}}=\frac{1\mathrm{~m}}{2\mathrm{~s}}=0.5\mathrm{~m/s}\)Answer: The average velocity between \(t=4\mathrm{~s}\) and \(t=6\mathrm{~s}\) is \(0.5\mathrm{~m/s}\)
