The position of a particle moving along x-axis varies with time ‘t’ given by $x (t) = 10 + 8 t - 3 t^{2}$ . Another particle is moving along the y-axis with its coordinate as a function of time given by y(t)= $5 - 8 t^{3}$ . At t=1 s, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{ν}$ . Then $ν$ in (m²/s²) is _______

Moderate

Solution

$\begin{array}{l} C o o r d i n a t e o f p a r t i c l e \\ A = X_{A} = - 3 t^{2} + 8 t + 10 \\ D i f f e r e n t i a t i n g w i t h t i m e w i l l g i v e v e l o c i t y o f p a r t i c l e A \\ {\vec{v}}_{A} = (- 6 t + 8) \overset{\land}{i} \\ A t t = 1 s, {\vec{v}}_{A} = 2 \overset{\land}{i} \\ C o o r d i n a t e o f p a r t i c l e \\ B = Y_{B} = 5 - 8 t^{3} \\ D i f f e r e n t i a t i n g w i t h t i m e w i l l g i v e v e l o c i t y o f p a r t i c l e B \\ {\vec{v}}_{B} = - 24 t^{2} \overset{\land}{j} \\ A t t = 1 s, {\vec{v}}_{B} = - 24 \overset{\land}{j} \\ S p e e d o f B w r t A = |{\vec{v}}_{B} - {\vec{v}}_{A}| = |- 24 \overset{\land}{j} - 2 \overset{\land}{i}| = \sqrt{24^{2} + 2^{2}} = \sqrt{580} m / s \end{array}$

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