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Byju's Answer

Standard XII

Physics

Position Time Graph

Two stones ar...

Question

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m $s^{- 1}$ and 30 m $s^{- 1}$ respectively. The time variation of the relative position of the second stone with respect to the first as shown in the figure. The equation of the liner part is

x_{2} - x_{1}

= 50t

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x_{2} - x_{1}

= 10t

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x_{2} - x_{1}

= 15t

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x_{2} - x_{1}

= 20t

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Solution

The correct option is C $x_{2} - x_{1}$ = 15t
Let $x = 0,$
$x_{1} = 200 + (15 t + \frac{1}{2} (- 10) t^{2})$
$x_{2} = 200 + (30 t + \frac{1}{2} (- 10) t^{2})$
$⟹ x_{2} - x_{1} = 15 t$

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Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first.
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Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s−1 and 30 m s−1 respectively. The time variation of the relative position of the second stone with respect to the first as shown in the figure. The equation of the liner part is

The correct option is C x2−x1 = 15tLet x=0,x1=200+(15t+12(−10)t2)x2=200+(30t+12(−10)t2)⟹x2−x1=15t

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m $s^{- 1}$ and 30 m $s^{- 1}$ respectively. The time variation of the relative position of the second stone with respect to the first as shown in the figure. The equation of the liner part is

The correct option is C $x_{2} - x_{1}$ = 15t
Let $x = 0,$
$x_{1} = 200 + (15 t + \frac{1}{2} (- 10) t^{2})$
$x_{2} = 200 + (30 t + \frac{1}{2} (- 10) t^{2})$
$⟹ x_{2} - x_{1} = 15 t$