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TS EAMCET · Physics · Motion In One Dimension

A particle is moving in xy-plane as \(\vec{x}=\left(4 t+t^2\right) \hat{i}, \vec{y}=\left(2 t+\frac{t^2}{2}\right) \hat{j}\) where \(\vec{x} \& \vec{y}\) are displacements measured along \(\mathrm{x}\) and y axes respectively in meters and \(t\) in seconds. What is the velocity of the particle?

  1. A \(\vec{v}=(4+t) \hat{i}+(2+t) \hat{j}\)
  2. B \(\vec{v}=(4+2 t) \hat{i}+(2+t) \hat{j}\)
  3. C \(\vec{v}=(4+2 t) \hat{i}+\left(2+\frac{t}{2}\right) \hat{j}\)
  4. D \(\vec{v}=(4+t) \hat{i}+\left(2+\frac{t}{2}\right) \hat{j}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\vec{v}=(4+2 t) \hat{i}+(2+t) \hat{j}\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}\) \[ =\frac{d \vec{x}}{d t}+\frac{d \vec{y}}{d t}=(4+2 t) \hat{i}+(2+t) \hat{j} \]
From TS EAMCET
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