A particle moves in a straight line and its speed depends on time as `v=|2t-3|` `int vdt` represent the distance travelled of the particle then find the displacement of the particle in `5 s`

Explanation

Given that,


`v=|2t-3|={(2t-3\ \ \ \ \ \ t ge (3)/(2)),(-2t+3 \ \ \ \ \ \ \ t lt (3)/(2)):} `


`S=int_0^5 vdt=int_0^(frac{3}{2}) vdt+int_(frac{3}{2})^5 vdt`


`S=int_0^5 vdt=int_0^(frac{3}{2}) (-2t+3)dt+int_(frac{3}{2})^5 (2t-3)dt`


`S=[-t^2+3t]_0^(3/2)+[t^2-3t]_(3/2)^5`


`S=-(3/2)^2+3*(3/2)+25-15-(3/2)^2+3*(3/2)`


`S=-2(9/4)+9+10`


`S=-9/2+19=29/2`


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