Let's break down the given problem and analyze the information provided in the acceleration-time graph.

Problem Analysis:

1. Graph Description: The acceleration-time graph shows that:

   • From $t = 0$ to $t = 3$ seconds, the car has a constant acceleration of $2 \, \text{m/s}^2$.
   • From $t = 3$ to $t = 5$ seconds, the car’s acceleration linearly increases from $2 \, \text{m/s}^2$ to $5 \, \text{m/s}^2$.
   • From $t = 5$ to $t = 6$ seconds, the acceleration decreases sharply from $5 \, \text{m/s}^2$ to $-5 \, \text{m/s}^2$.
   • From $t = 6$ to $t = 7$ seconds, the car maintains a constant acceleration of $-5 \, \text{m/s}^2$.

2. Initial Speed: When the car passes point $P$ (at $t = 0$), its speed is $12 \, \text{m/s}$.

3. Questions:

   • (i) Explain why the speed is greatest at $t = 6$ seconds.
   • (ii) Calculate the speed at $t = 3$ seconds and the distance traveled from $P$ up to this time.

Solution:

Part (i) Explanation:

The speed is the greatest at $t = 6$ seconds because the car has been accelerating in the positive direction up until this point, increasing its velocity. At $t = 6$ seconds, the car reaches the maximum positive acceleration (the peak area under the curve), after which the acceleration becomes negative, causing the speed to decrease. Thus, $t = 6$ is the point of maximum speed.

Part (ii) Calculations:

1. Speed at $t = 3$ seconds:

   • From $t = 0$ to $t = 3$, the acceleration is constant at $2 \, \text{m/s}^2$.
   • Using the formula $v = u + at$, where:
     • $u = 12 \, \text{m/s}$ (initial speed),
     • $a = 2 \, \text{m/s}^2$ (acceleration),
     • $t = 3$ seconds.

   $v = 12 + (2 \times 3) = 12 + 6 = 18 \, \text{m/s}$ So, the speed at $t = 3$ seconds is $18 \, \text{m/s}$.

2. Distance from $P$ at $t = 3$ seconds:

   • Using the formula $s = ut + \frac{1}{2}at^2$, where:
     • $u = 12 \, \text{m/s}$,
     • $a = 2 \, \text{m/s}^2$,
     • $t = 3$ seconds.

   $s = (12 \times 3) + \frac{1}{2} \times 2 \times (3)^2$ $s = 36 + \frac{1}{2} \times 2 \times 9$ $s = 36 + 9 = 45 \, \text{meters}$

   So, the distance from $P$ at $t = 3$ seconds is $45 \, \text{meters}$.

Would you like further details on any step or have additional questions?

Related Questions:

1. What is the speed of the car at $t = 6$ seconds?
2. How would the graph look if the car decelerated from the start?
3. What is the total distance traveled by the car in the 7 seconds shown on the graph?
4. How does varying acceleration affect the total distance traveled?
5. How would the velocity-time graph look for the same scenario?

Tip:

When working with acceleration-time graphs, remember that the area under the curve represents the change in velocity over time.