A body is moving with a uniform speed of $20 \mathrm{~ms}^{-1}$ on a horizontal circle. The change in…
- $20 \mathrm{~ms}^{-1}$
- $10 \mathrm{~ms}^{-1}$
- $40 \mathrm{~ms}^{-1}$
- $\frac{20}{\sqrt{2}} \mathrm{~ms}^{-1}$
Solution

$\begin{aligned} & \Delta v=\vec{v}_2-\vec{v}_1=-v \hat{j}-v \hat{j}=-2 v \hat{j} \\ & |\Delta v|=2 v=2 \times 20=40 \mathrm{~m} / \mathrm{s}\end{aligned}$
Asked in: AP EAMCET 2023 (17 May Shift 2)
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