Mathematics 9709/42 — October/November 2021

Walkthrough

The problem states that the two accelerating stages have equal acceleration. Acceleration on a velocity-time graph is given by the gradient of the line segment.

First, we find the acceleration during the initial stage from $t=0$ to $t=5$. The velocity increases from $0$ to $20 \text{ m s}^{-1}$, so the gradient is $\frac{20 - 0}{5 - 0} = 4 \text{ m s}^{-2}$.

Next, we find the acceleration during the second stage from $t=T$ to $t=50$. The velocity increases from $6$ to $20 \text{ m s}^{-1}$ over a time of $(50 - T)$ seconds, giving a gradient of $\frac{20 - 6}{50 - T} = \frac{14}{50 - T}$.

Setting these two gradients equal gives $\frac{14}{50 - T} = 4$. Cross-multiplying yields $14 = 200 - 4T$, which rearranges to $4T = 186$, so $T = 46.5$.

Key Takeaways

• The gradient of a velocity-time graph represents acceleration.
• Equating gradients from different segments allows you to solve for unknown time values.

Common Mistakes

• Using the wrong time interval for the second acceleration (e.g., using $50$ instead of $50 - T$).
• Forgetting that the velocity at $t=T$ is $6 \text{ m s}^{-1}$, not $0$.

Things to Be Careful About

• Ensure the time interval is calculated correctly as the difference between the end and start times ($50 - T$).
• Check that the algebraic manipulation of the fraction is correct when cross-multiplying.

Walkthrough

Distance is the area under a velocity-time graph. The graph is composed of six distinct segments, which create six geometric regions between the graph and the time axis. We calculate the area of each region individually and add them together.

From $t=0$ to $t=5$, the shape is a triangle with base $5$ and height $20$, giving area $50$.
From $t=5$ to $t=25$, the shape is a rectangle with width $20$ and height $20$, giving area $400$.
From $t=25$ to $t=30$, the shape is a trapezium with parallel sides $20$ and $6$, and width $5$, giving area $65$.
From $t=30$ to $t=T=46.5$, the shape is a rectangle with width $16.5$ and height $6$, giving area $99$.
From $t=T=46.5$ to $t=50$, the shape is a trapezium with parallel sides $6$ and $20$, and width $3.5$, giving area $45.5$.
From $t=50$ to $t=60$, the shape is a triangle with base $10$ and height $20$, giving area $100$.

Summing these areas gives the total distance: $50 + 400 + 65 + 99 + 45.5 + 100 = 759.5$ metres.

Key Takeaways

• The area under a velocity-time graph represents the total distance travelled.
• Complex shapes under the graph can be decomposed into simple geometric figures (triangles, rectangles, trapeziums) for easier calculation.

Common Mistakes

• Forgetting to substitute the value of $T$ found in part (a) before calculating the areas of the last two regions.
• Using the wrong formula for the area of a trapezium (must use $\frac{1}{2}(a+b)h$, not $ab$).

Things to Be Careful About

• Ensure all time intervals are calculated as differences (e.g., $T - 30$ and $50 - T$).
• Check arithmetic when summing the final areas to avoid simple addition errors.