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Relative Velocity

Positions of ...

Question

Positions of two particles $A$ and $B$ as a function of time are given by $X_{A} = - 3 t^{2} + 8 t + c$ and $Y_{B} = 10 - 8 t^{3}$ . The velocity of $B$ with respect to $A$ at $t = 1$ is $\sqrt{v}$ . Find $v$ [in $m^{2} / s^{2}$ ]

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Solution

As we know,
$\frac{d x}{d t} = v$
$X_{A} = - 3 t^{2} + 8 t + c$
$\Rightarrow \frac{d {\to X}_{A}}{d t} = {\to v}_{A} = (- 6 t + 8)^i$
$\Rightarrow {\to v}_{A} (t = 1 s) = 2^i m/s$

$Y_{B} = 10 - 8 t^{3}$
$\Rightarrow \frac{d {\to Y}_{B}}{d t} = {\to v}_{B} = - 24 t^{2}^j$
$\Rightarrow {\to v}_{B} (t = 1 s) = - 24^j m/s$
Therefore,
$| {\to v}_{B, A} | = ∣ ∣ {\to v}_{B} - {\to v}_{A} ∣ ∣ = ∣ ∣ - 24^j - 2^i ∣ ∣$
$= \sqrt{24^{2} + 2^{2}} = \sqrt{580} m/s$
$\Rightarrow ∣ ∣ {\to v}_{B} - {\to v}_{A} ∣ ∣ = \sqrt{v} = \sqrt{580}$ [as given]
So, $v = 580 m^{2} / s^{2}$

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Relative Velocity

Standard XII Physics

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