Answer:
Area of the triangle gives the total distance travelled by the particle.
Area of\[\Delta OAB=\frac{1}{2}\times 205=50m\]
Area of\[\Delta BCD=\frac{1}{2}\times 20\times 5=50m\]
\[{{t}_{2}}=\frac{{{s}_{2}}}{{{V}_{2}}}=\frac{s/4}{20\,km/h}=\frac{s}{80}hr\]total distance travelled by the particle
\[=50m+50m=100m.\]
Average velocity\[=\frac{total\,\,displacement}{total\,\,time}\]
Total displacement travelled by the particle
\[=50-50=zero.\]Hence, average velocity during this period = zero.
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