Five persons A, B, C, D & E are pulling a cart of mass 100kg on a smooth surface and cart is moving with acceleration $3 m / s^{2}$ in east direction. When person ‘A’ stops pulling, it moves with acceleration $1 m / s^{2}$ in the west direction. When person ‘B’ stops pulling, it moves with acceleration $24 m / s^{2}$ in the north direction. The magnitude of acceleration of the cart when only A & B pull the cart keeping their directions same as the old direction is

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$26 m / s^{2}$

$3 \sqrt{71} m / s^{2}$

$25 m / s^{2}$

$30 m / s^{2}$

answer is C.

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Detailed Solution

${\bar{F}}_{A} + {\bar{F}}_{B} + {\bar{F}}_{C} + {\bar{F}}_{D} + {\bar{F}}_{E} = (3 \hat{i}) 100 = 300 \hat{i}$
${\bar{F}}_{B} + {\bar{F}}_{C} + {\bar{F}}_{D} + {\bar{F}}_{E} = 1 (- \hat{i}) 100 = - 100 \hat{i}$
${\bar{F}}_{A} + {\bar{F}}_{C} + {\bar{F}}_{D} + {\bar{F}}_{E} = 24 (- \hat{j}) 100 = 2400 \hat{j}$
(1) & (2) ${\bar{F}}_{A} = 400 \hat{i} {\bar{F}}_{B} = 300 \hat{i} - 2400 \hat{j}$
${\bar{F}}_{A} + {\bar{F}}_{B} = 700 \hat{i} - 2400 \hat{j}$
$\vec{a} = \frac{700 \hat{i} - 2400 \hat{j}}{100} = 7 \hat{i} - 24 \hat{j}$
$| \vec{a} | = \sqrt{7^{2} + {(24)}^{2}} = \sqrt{49 + 576} = \sqrt{625}$
$a = 25 m / s^{2}$