1. **Problem Statement:** Given a velocity vs time graph, we need to: a. Draw the acceleration vs time graph. b. Determine the position at given times using the velocity data. c. Use the position data to draw the position vs time graph. 2. **Key formulas and concepts:** - Acceleration $a(t)$ is the slope of the velocity vs time graph: $$a(t) = \frac{dv}{dt}$$ - Position $x(t)$ is found by integrating velocity over time: $$x(t) = x(t_0) + \int_{t_0}^t v(t) dt$$ - Given initial position values at certain times, we can calculate position changes by finding the area under the velocity curve between times. 3. **Step a: Draw acceleration vs time graph** - Find slopes of velocity segments: - From $t=0$ to $t=3$: velocity drops from $-2$ to $-3$, slope = $\frac{-3 - (-2)}{3 - 0} = \frac{-1}{3} = -\frac{1}{3}$ m/s² - From $t=3$ to $t=7$: velocity rises from $-3$ to $3$, slope = $\frac{3 - (-3)}{7 - 3} = \frac{6}{4} = 1.5$ m/s² - From $t=7$ to $t=10$: velocity constant at $3$, slope = $0$ - From $t=10$ to $t=14$: velocity falls from $3$ to $-2$, slope = $\frac{-2 - 3}{14 - 10} = \frac{-5}{4} = -1.25$ m/s² - The acceleration graph is piecewise constant with these values over the respective intervals. 4. **Step b: Determine position at given times** - Given positions: - $x(2) = -3$ m - $x(3) = -4.5$ m - $x(4) = -5$ m - Calculate position at $t=5$ and $t=6$ by integrating velocity from known points. - From $t=3$ to $t=4$: - Velocity changes linearly from $-3$ to about $0$ (since at $t=7$ velocity is 3, slope is 1.5, so at $t=4$, velocity = $-3 + 1.5 \times (4-3) = -3 + 1.5 = -1.5$ m/s) - Average velocity between 3 and 4: $\frac{-3 + (-1.5)}{2} = -2.25$ m/s - Position change: $\Delta x = v_{avg} \times \Delta t = -2.25 \times 1 = -2.25$ m - So, $x(4) = x(3) + \Delta x = -4.5 + (-2.25) = -6.75$ m (But given is -5 m, so we use given data) - From $t=4$ to $t=5$: - Velocity at $t=5$: $v(5) = -3 + 1.5 \times (5-3) = -3 + 3 = 0$ m/s - Average velocity between 4 and 5: $\frac{-1.5 + 0}{2} = -0.75$ m/s - Position change: $\Delta x = -0.75 \times 1 = -0.75$ m - Position at 5: $x(5) = x(4) + \Delta x = -5 + (-0.75) = -5.75$ m - From $t=5$ to $t=6$: - Velocity at $t=6$: $v(6) = -3 + 1.5 \times (6-3) = -3 + 4.5 = 1.5$ m/s - Average velocity between 5 and 6: $\frac{0 + 1.5}{2} = 0.75$ m/s - Position change: $\Delta x = 0.75 \times 1 = 0.75$ m - Position at 6: $x(6) = x(5) + \Delta x = -5.75 + 0.75 = -5$ m 5. **Step c: Draw position vs time graph** - Use the table: | Time (s) | Position (m) | |----------|--------------| | 2 | -3 | | 3 | -4.5 | | 4 | -5 | | 5 | -5.75 | | 6 | -5 | - Plot these points and connect smoothly to show position changes over time. **Final answers:** - Acceleration values piecewise: $a(t) = \begin{cases} -\frac{1}{3}, & 0 \leq t < 3 \\ 1.5, & 3 \leq t < 7 \\ 0, & 7 \leq t < 10 \\ -1.25, & 10 \leq t \leq 14 \end{cases}$ - Positions: - $x(5) = -5.75$ m - $x(6) = -5$ m