Answer:
(a) Area under triangle give the total distance travelled by the particle.
Area of \[\Delta OAB=\frac{1}{2}\times 20\times 5=50m\]
Area of \[\Delta BCD=\frac{1}{2}\times 20\times 5=50m\]
\[24=u+\frac{a}{2}(2\times 10-1)\] total distance travelled by the particle \[=50\text{ }m+50\text{ }m=100\text{ }m.\]
(b) total displacement travelled by the particle\[=50-50=\]zero. Hence, average velocity during this period = zero.
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