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Grade 11Mechanics

Please tell me properly about distance calculation .......
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Profile image of Parnav Bansal
8 Years agoGrade 11
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1 Answer

Profile image of Eshan
ApprovedApproved Tutor Answer8 Years ago
Dear student,

In this motion, intially particle starts moving in negative direction, stop at t=2s, and reverses its direction.

The displacement of particle is given ass(t)=\int_0^t(3t^2-6t)dt=t^3-3t^2

Hence the distance travelled between t=0s to t=2s iss_1=\left|(2)^3-3(2)^2\right|=4m

Distance travelled from t=2s to t=3.5s=s_2=(3.5)^3-3(3.5)^2-(2)^3+3(2)^2=10.12m

Hence the total distance travelled by body=s_1+s_2=14.12m