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
A1 = 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
A2 = Area of triangle
$\displaystyle A_2 = \frac{1}{2} \times 1 \times 10 $
A2 = 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