A train starts from rest with uniform acceleration for 20 seconds. It travels with uniform velocity for another 30 seconds the brakes are applied and then the train comes to rest in a further 10 seconds Sketch the velocity time graph and use it to calculate the total distance covered by the train

The velocity-time graph will consist of three segments:

• From $t=0$ s to $t=20$ s: The train starts from rest ($v=0$) and accelerates uniformly. This is represented by a straight line segment starting from the origin $(0,0)$ and rising to a maximum velocity, let's call it $V$, at $t=20$ s. So, the first point is $(0,0)$ and the second point is $(20, V)$.
• From $t=20$ s to $t=50$ s: The train travels with uniform velocity $V$. This is represented by a horizontal straight line segment from $(20, V)$ to $(50, V)$.
• From $t=50$ s to $t=60$ s: The brakes are applied, and the train decelerates uniformly to rest ($v=0$) in 10 seconds. This is represented by a straight line segment from $(50, V)$ down to $(60, 0)$.

The overall shape of the graph is a trapezoid.

The total distance covered by the train is equal to the area under the velocity-time graph. This area can be divided into three parts: a triangle (acceleration), a rectangle (uniform velocity), and another triangle (deceleration).

Area of the first triangle (acceleration phase):

$$A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (20 \text{ s}) \times V$$ $$A_1 = 10V$$

Area of the rectangle (uniform velocity phase):

$$A_2 = \text{base} \times \text{height} = (50 \text{ s} - 20 \text{ s}) \times V = (30 \text{ s}) \times V$$ $$A_2 = 30V$$

Area of the second triangle (deceleration phase):

$$A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (60 \text{ s} - 50 \text{ s}) \times V = \frac{1}{2} \times (10 \text{ s}) \times V$$ $$A_3 = 5V$$

Total distance $D = A_1 + A_2 + A_3$

$$D = 10V + 30V + 5V$$ $$D = 45V$$

Since the maximum velocity $V$ is not given, the total distance is expressed in terms of $V$.

The total distance covered by the train is $\text{45V units}$.