A car starts from rest and accelerates uniformly to a speed of 40 m/s in 1 min 40 s. It then travels at a constant speed for 5 minutes before decelerating uniformly to rest in 2 minutes. a) Draw a vel

Here's the solution to question 9:

The car's journey can be broken down into three phases:

Phase 1: Acceleration

• Starts from rest ($0 \text{ m/s}$).
• Reaches $40 \text{ m/s}$.
• Time taken: $1 \text{ min } 40 \text{ s} = 60 \text{ s} + 40 \text{ s} = 100 \text{ s}$.
• End of phase 1: $(100 \text{ s}, 40 \text{ m/s})$.

Phase 2: Constant speed

• Speed: $40 \text{ m/s}$.
• Time taken: $5 \text{ min} = 5 \times 60 \text{ s} = 300 \text{ s}$.
• This phase starts at $100 \text{ s}$ and ends at $100 \text{ s} + 300 \text{ s} = 400 \text{ s}$.
• End of phase 2: $(400 \text{ s}, 40 \text{ m/s})$.

Phase 3: Deceleration

• Starts at $40 \text{ m/s}$.
• Comes to a halt ($0 \text{ m/s}$).
• Time taken: $2 \text{ min} = 2 \times 60 \text{ s} = 120 \text{ s}$.
• This phase starts at $400 \text{ s}$ and ends at $400 \text{ s} + 120 \text{ s} = 520 \text{ s}$.
• End of phase 3: $(520 \text{ s}, 0 \text{ m/s})$.

a) Draw a velocity-time graph to represent the information above.

To draw the graph:

1. Draw a horizontal axis for time (in seconds) and a vertical axis for velocity (in m/s).
2. Plot the following points:
   • $(0, 0)$ (start from rest)
   • $(100, 40)$ (end of acceleration)
   • $(400, 40)$ (end of constant speed)
   • $(520, 0)$ (car comes to a halt)
3. Connect $(0,0)$ to $(100,40)$ with a straight line (acceleration).
4. Connect $(100,40)$ to $(400,40)$ with a horizontal line (constant velocity).
5. Connect $(400,40)$ to $(520,0)$ with a straight line (deceleration). The resulting graph will be a trapezium.

b) Use your graph to find:

(i) The initial acceleration.
The initial acceleration is the gradient of the first segment of the graph (from $t=0 \text{ s}$ to $t=100 \text{ s}$).

$$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Change in time}}$$ $$\text{Acceleration} = \frac{40 \text{ m/s} - 0 \text{ m/s}}{100 \text{ s} - 0 \text{ s}}$$ $$\text{Acceleration} = \frac{40}{100} \text{ m/s}^2$$ $$\text{Acceleration} = 0.4 \text{ m/s}^2$$

The initial acceleration is $\boxed{\text{0.4 m/s}^2}$.

(ii) The deceleration when the car is brought to a halt.
The deceleration is the magnitude of the gradient of the last segment of the graph (from $t=400 \text{ s}$ to $t=520 \text{ s}$).

$$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Change in time}}$$ $$\text{Acceleration} = \frac{0 \text{ m/s} - 40 \text{ m/s}}{520 \text{ s} - 400 \text{ s}}$$ $$\text{Acceleration} = \frac{-40 \text{ m/s}}{120 \text{ s}}$$ $$\text{Acceleration} = -\frac{1}{3} \text{ m/s}^2$$

The deceleration is the positive value of this acceleration.

$$\text{Deceleration} = \frac{1}{3} \text{ m/s}^2 \approx 0.333 \text{ m/s}^2$$

The deceleration is $\boxed{\text{0.333 m/s}^2}$ (to 3 significant figures).

(iii) The distance traveled in km.
The total distance traveled is the area under the entire velocity-time graph (a trapezium).
The formula for the area of a trapezium is $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.

• Height of the trapezium (maximum velocity) = $40 \text{ m/s}$.
• Length of the top parallel side (duration of constant velocity) = $400 \text{ s} - 100 \text{ s} = 300 \text{ s}$.
• Length of the bottom parallel side (total time) = $520 \text{ s}$.

$$\text{Total distance} = \frac{1}{2} \times (300 \text{ s} + 520 \text{ s}) \times 40 \text{ m/s}$$ $$\text{Total distance} = \frac{1}{2} \times 820 \text{ s} \times 40 \text{ m/s}$$ $$\text{Total distance} = 410 \text{ s} \times 40 \text{ m/s}$$ $$\text{Total distance} = 16400 \text{ m}$$

To convert meters to kilometers:

$$\text{Distance in km} = \frac{16400 \text{ m}}{1000 \text{ m/km}}$$ $$\text{Distance in km} = 16.4 \text{ km}$$

The total distance traveled is $\boxed{\text{16.4 km}}$.