Question 127: Motion in 1D

Question

The position of a particle is given by
$$\vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}
$$
where $$\mathrm{t}$$ is in seconds and $$\mathrm{r}$$ in metre. Find the magnitude and direction of velocity $$v(t)$$, at $$t=1 \mathrm{~s}$$, with respect to $$\mathrm{x}$$-axis

Accepted Answer

$$\begin{aligned} & \overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \ & 	ext { at } \mathrm{t}=1 \mathrm{sec} \ & \overrightarrow{\mathrm{V}}=4 \hat{\mathrm{i}}+4(1) \hat{\mathrm{j}} \ & |\overrightarrow{\mathrm{V}}|=\sqrt{4^2+4^2}=4 \sqrt{2} \ & 	an \alpha=\frac{4}{4}=1 \ & \alpha=45^{\circ}\end{aligned}$$

Answer

$$4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$$

Answer

$$4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ}$$

Answer

$$3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ}$$

Answer

$$3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$$

Motion in 1D

The position of a particle is given by $\vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k} where \mathrm{t} is in seconds and \mathrm{r} in metre. Find the magnitude and direction of velocity v(t), at t=1 \mathrm{~s}, with respect to \mathrm{x}$-axis

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