The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by

airplane in the first 30.5 second is _______ km .

Let us divide this problem into 2 parts:-

First 2 seconds where the velocity changes , which means acceleration is present

Second part from 2 seconds to 30.5 (28.5 seconds in total) seconds where Velocity is constant

We know that distance s:-

$$s=ut+\ \frac{\ 1}{2}at^2$$

For first 2 seconds :-

u=$$200\ \ \frac{\ m}{s}$$

a=$$\ \frac{\ v_f=v_i}{t}$$

a=$$\ \frac{\ 400-200}{2}$$

a=100 $$\ \frac{\ m}{s^2}$$

Now $$s=200\times\ 2\ +\ \ \frac{\ 1}{2}\times\ 100\times\ \left(2\right)^2$$

$$s=600\ m$$

For next 28.5 s a=0(constant velocity), Here u=400 $$\ \frac{\ m}{s}$$

so s=$$ut$$

s=$$400\times\ 28.5$$

s=11400 m

now adding both 

$$s_{total}$$ = 600+ 11400

$$s_{total}$$= 12000 m = 12 Km