A bus travels from rest with uniform acceleration for 30 s, then at a constant speed for 60 s, and finally decelerates uniformly to rest in 40 s. The total distance covered is 2090 m. Calculate (a) th

The graph is a trapezium. The area of a trapezium is given by $A = \frac{1}{2}(a+b)h$, where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height.
In this case, the parallel sides are the time intervals where speed is constant (top) and the total time (bottom). The height is the maximum speed ($V_{max}$).
The duration of constant speed is from $t=30$ s to $t=90$ s, so $a = 90 - 30 = 60$ s.
The total time is from $t=0$ s to $t=130$ s, so $b = 130 - 0 = 130$ s.
The total distance is given as 2090 m.

To convert m/s to km/h, multiply by $\frac{3600}{1000}$ or 3.6.

The maximum speed attained is $\boxed{\text{79.2 km/h}}$.

The bus starts from rest, so initial speed $u = 0$ m/s.
The final speed at $t=30$ s is the maximum speed, $v = V_{max} = 22$ m/s.
The time taken is $t = 30$ s.

Acceleration is the gradient of the speed-time graph, given by $a = \frac{\text{change in speed}}{\text{change in time}}$.

The acceleration in the first 30 seconds is $\boxed{\frac{11}{15} \text{ m/s}^2 \text{ or approximately } 0.733 \text{ m/s}^2}$.

The total journey time is 130 seconds. The last 20 seconds are from $t = 130 - 20 = 110$ s to $t = 130$ s.

The deceleration phase is from $t=90$ s to $t=130$ s.
Initial speed at $t=90$ s is $V_{max} = 22$ m/s.
Final speed at $t=130$ s is 0 m/s.
Time taken for deceleration is $130 - 90 = 40$ s.

So, the acceleration is $-0.55$ m/s$^2$.

Step