1. **Problem statement:** We are given the velocity function of a particle moving along a straight line: $$v = 3t(t - 4)$$ for time $$0 \leq t \leq k$$. We need to sketch the velocity-time graph and calculate the total distance travelled during this time. 2. **Sketching the velocity-time graph:** - The velocity function is a quadratic: $$v = 3t^2 - 12t$$. - It is zero at $$t=0$$ and $$t=4$$. - For $$0 < t < 4$$, velocity changes from 0 to a maximum and back to 0. - For $$t > 4$$, velocity becomes negative (particle moves left). 3. **Finding critical points:** - Differentiate velocity to find maxima/minima: $$\frac{dv}{dt} = 6t - 12$$ - Set derivative to zero: $$6t - 12 = 0 \Rightarrow t = 2$$ - Velocity at $$t=2$$: $$v(2) = 3 \times 2 \times (2-4) = 3 \times 2 \times (-2) = -12$$ 4. **Velocity-time graph shape:** - At $$t=0$$, $$v=0$$. - At $$t=2$$, $$v=-12$$ (minimum velocity). - At $$t=4$$, $$v=0$$. - The graph is a parabola opening upwards with vertex at $$t=2$$. 5. **Calculating total distance travelled from $$t=0$$ to $$t=k$$:** - Total distance is the integral of the absolute value of velocity: $$\text{Distance} = \int_0^k |v(t)| dt$$ - Since velocity changes sign at $$t=0$$ and $$t=4$$, split the integral accordingly. - For $$0 \leq t \leq 4$$, velocity is negative (since vertex is minimum at -12), so: $$|v(t)| = -v(t) = -3t(t-4) = 12t - 3t^2$$ - For $$t > 4$$, velocity is positive, so $$|v(t)| = v(t)$$. 6. **Distance travelled for $$0 \leq t \leq 4$$:** $$\int_0^4 (12t - 3t^2) dt = \left[6t^2 - t^3\right]_0^4 = (6 \times 16 - 64) - 0 = (96 - 64) = 32$$ 7. **Distance travelled for $$4 < t \leq k$$:** $$\int_4^k 3t(t-4) dt = \int_4^k (3t^2 - 12t) dt = \left[t^3 - 6t^2\right]_4^k = (k^3 - 6k^2) - (64 - 96) = k^3 - 6k^2 - (-32) = k^3 - 6k^2 + 32$$ 8. **Total distance travelled from $$0$$ to $$k$$:** - If $$k \leq 4$$, distance is: $$\int_0^k |v(t)| dt = \int_0^k (12t - 3t^2) dt = \left[6t^2 - t^3\right]_0^k = 6k^2 - k^3$$ - If $$k > 4$$, distance is: $$32 + (k^3 - 6k^2 + 32) = k^3 - 6k^2 + 64$$ --- **Final answers:** - Velocity-time graph is a parabola with zeros at $$t=0$$ and $$t=4$$, minimum velocity $$-12$$ at $$t=2$$. - Total distance travelled from $$0$$ to $$k$$ is: $$\text{Distance} = \begin{cases} 6k^2 - k^3, & 0 \leq k \leq 4 \\ k^3 - 6k^2 + 64, & k > 4 \end{cases}$$