question_answer

The particle moves according to given velocity-time graph. The ratio of travelling distance in last 2s and 7s will be

A)  \[\frac{1}{4}\]

B)  \[\frac{1}{2}\]

C)  \[\frac{1}{8}\]

D)  \[\frac{1}{6}\]

Correct Answer: B

Solution :

 Travelling distance in last 2 second = Area of C \[=\frac{1}{2}\times 2\times 10=10\,m\] Total distance in 7 s = Area of A + Area of B + Area of C \[=\frac{1}{2}\times 2\times 10+2\times 10+\frac{1}{2}\times 2\times 10\] \[=40m\] The ratio of distances\[=\frac{10}{40}=\frac{1}{4}\]