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2 years ago

(4j3 - 7j + 2s3 ) - ( - 12j3 - 8j )

Mathematics

1 answer:

OLga [1]2 years ago

7 0

Answer:

$-8j^3+j+2s^3$

Step-by-step explanation:

The trick here is to consider $j^3$ , $j$ , and $s^3$ as separate variables.

$(4j^3-7j+2s^3)-(-12j^3-8j)$

$4j^3-7j+2s^3+12j^3+8j$

$4j^3+12j^3-7j+8j+2s^3$

$16j^3+j+2s^3$

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2 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

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Given that $a= \frac{dv}{dt} =-kv^2$

Then,

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$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

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For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

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Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

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