The velocity-time graph of a body moving in a straight line is shown in Fig. The displacement of the body in…

The velocity-time graph of a body moving in a straight line is shown in Fig. The displacement of the body in \(10 \mathrm{~s}\) is
  1. \(4 \mathrm{~m}\)
  2. \(6 \mathrm{~m}\)
  3. \(8 \mathrm{~m}\)
  4. \(10 \mathrm{~m}\)

Solution

We know that area under \(v-t\) graph is displacement.
Area from 0 to \(6 \mathrm{~s}=\frac{1}{2} \times 6 \times 2=6 \mathrm{~m}\)
Area from 6 to \(8 s=\frac{1}{2} \times 2 \times(-2)=-2 m\)
Area from 8 to \(10 \mathrm{~s}=2 \times 1=2 \mathrm{~m}\)
So Net displacement \(=6-2+2=6 \mathrm{~m}\)

Asked in: JEE Mains - Motion In One Dimension - Chapter Test

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