question_answer

From the velocity - time plot shown in figure, (a) what is the distance travelled by the particle during the first 40 seconds. (b) The average velocity during this period is ______

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.