KinematicsMedium
Question
The velocity( $v$ ) versus time $(t)$ plot of a particle is shown in the figure, for a time interval of 40s. The total distance travelled by the particle and the average velocity during this period are, respectively $\_\_\_\_$
Options
A.25 m and zero
B.50 m and zero
C.100 m and zero
D.100 m and $2.5\text{ }m/s$
Solution
Sol. [Key Concept: Distance $=$ area under $|v|$–$t$; average velocity $= \frac{net\ displacement}{time} = 0$.]
Total distance travel
$${d = \frac{1}{2}(20)(5) + \frac{1}{2}(20) \times 5 }{d = 100\text{ }m }$$Average velocity $= \frac{\text{~}\text{Total displacement}\text{~}}{\text{~}\text{Total time}\text{~}}$
$$< v > = \frac{0}{40} = 0 $$
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