First slide

Instantaneous velocity

Question

The position vector of a particle moving in xy plane is given by $\vec{r} = (t^{2} - 4) \hat{i} + (t - 4) \hat{j}$ . Find velocity when particle will cross y-axis

Easy

A

$\sqrt{17} m / s$
B

4 m / s
C

2 m / s
D

1 m / s

Solution

when the particle crosses Y axis, x=0
$\begin{array}{l} t^{2} - 4 = 0 \Rightarrow t = 2 \sec \\ t = 2 s e c \\ V_{x} = \frac{d x}{d t} = 2 t = 4 m / s \\ V_{y} = \frac{d y}{d x} = 1 \\ V = \sqrt{4^{2} + 1^{2}} = \sqrt{17} \end{array}$

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