Solution 9
We are given the graph of average velocity $v_{av}$ versus time $t$ and need to derive the graph of instantaneous velocity $v$ versus time $t$.
Key Relationship: Since $v_{av} = \frac{S}{t}$, the displacement is $S(t) = t \cdot v_{av}(t)$. The instantaneous velocity is the derivative $v(t) = \frac{dS}{dt}$.
Interval 1: $0 < t < 1$ s
From the given graph, $v_{av}$ increases linearly from 0 to 2.
- Equation: $v_{av} = 2t$.
- Displacement: $S = t \cdot (2t) = 2t^2$.
- Velocity: $v = \frac{d}{dt}(2t^2) = 4t$.
At $t=1$, $v = 4 \, \text{m/s}$.
Interval 2: $1 < t < 2$ s
From the given graph, $v_{av}$ is constant at 2.
- Equation: $v_{av} = 2$.
- Displacement: $S = t \cdot 2 = 2t$.
- Velocity: $v = \frac{d}{dt}(2t) = 2 \, \text{m/s}$.
The velocity drops instantaneously from 4 to 2 at $t=1$ and remains constant.
Interval 3: $2 < t < 4$ s
From the given graph, $v_{av}$ increases linearly from 2 to 3.
- Slope $m = \frac{3-2}{4-2} = 0.5$.
- Equation: $v_{av} – 2 = 0.5(t – 2) \implies v_{av} = 0.5t + 1$.
- Displacement: $S = t(0.5t + 1) = 0.5t^2 + t$.
- Velocity: $v = \frac{d}{dt}(0.5t^2 + t) = t + 1$.
At $t=2$, $v = 2+1 = 3 \, \text{m/s}$. (Jump from 2 to 3)
At $t=4$, $v = 4+1 = 5 \, \text{m/s}$.
Figure 5: The derived instantaneous velocity vs. time graph.