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Q.

The position of a point in time ‘t’ is given by $x = 1 + 2 t + 3 t^{2}$ and $y = 2 - 3 t + 4 t^{2} .$ Then its acceleration at time ‘t’ is

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a

13

b

12

c

10

d

15

answer is A.

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Detailed Solution

$Given x = 1 + 2 t + 3 t^{2}; y = 2 - 3 t + 4 t^{2} \Rightarrow \frac{d x}{d t} = 2 + 6 t$

$; \frac{d y}{d t} = - 3 + 8 t \Rightarrow \frac{d^{2} x}{d t^{2}} = 6; \frac{d^{2} y}{d t^{2}} = 8$

$∴ acceleration = \sqrt{{(\frac{d^{2} x}{d t^{2}})}^{2} + {(\frac{d^{2} y}{d t^{2}})}^{2}}$

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