Question 25: Motion in 2D & Vectors

Question

The position vector of a moving body at any instant of time is given as $\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 s$ is,

Accepted Answer

$$\begin{aligned}
& \overrightarrow{\mathrm{r}}=5 \mathrm{t}^2 \hat{\mathrm{i}}-5 \hat{\mathrm{t}} \
& \overrightarrow{\mathrm{v}}=10 \mathrm{t} \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \
& \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad 	ext { at } \mathrm{t}=2 \mathrm{sec}
\end{aligned}$$
$$\begin{aligned}
& 	an 	heta=\frac{20}{5}=4 \
& 	heta=	an ^{-1} 4 \
& 	ext { From-veY-axis }
\end{aligned}$$

Answer

$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $	an ^{-1} 4$ with - ve Y axis

Answer

$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $	an ^{-1} 4$ with + ve $X$ axis

Answer

$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $	an ^{-1} 4$ with + ve $X$ axis

Answer

$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $	an ^{-1} 4$ with $-$ ve $Y$ axis

Motion in 2D & Vectors

The position vector of a moving body at any instant of time is given as \overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}. The magnitude and direction of velocity at t=2 s is,

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