Q: A particle moving on straight line whose velocity-time graph is shown in the figure. The average speed from t = 0 to t = 6 s is v = 15/n m/s. Find the value of n.

**Solution :** Area of v-t graph from zero to 5 sec is

A_{1} = Area of trapezium

$\displaystyle A_1 = \frac{1}{2}(5+3) \times 10 $

$\displaystyle A_1 = 40 m $

Area of v-t graph from 5 sec to 6 sec is

A_{2} = Area of triangle

$\displaystyle A_2 = \frac{1}{2} \times 1 \times 10 $

A_{2} = 5 m

Total distance traveled = 40 + 5 = 45 m

Average Speed = Total Distance / Total time

$\displaystyle V = \frac{45}{6} = \frac{15}{2}$

On comparing we get ,

**n = 2 **