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Two rockets are fired vertically into the air from the ground. The second rocket is launched four seconds after the first. The velocity of the first rocket is $v_1(t) = 6 − t$ metres per second and the velocity of the second is $v_2(t) = 10 − t$ metres per second,where $t$ is the time in seconds after the first launch.

How long after first the launch will both rockets be at the same height?

Below is my wrong solution (but I have no idea why it is wrong and that is my question QAQ) $$ \text{Let $x$ be the time elapsed after the first launch.} \\ \text{Then} \int_0^x v_1(t)dt = \int_0^{x-4}v_2(t)dt \\ \implies \int_0^x (6-t)dt = \int_0^{x-4}(10-t)dt \\ \implies (6t-\frac{1}{2}t^2)\Bigg |_0^x = (10t-\frac{1}{2}t^2)\Bigg | _0 ^{x-4} \\ \implies 6x = 10x -40 - 8 + 4x \implies x = 6 $$ However, the correct answer is $x=8$. Could anyone tell why the method I used is wrong?

Here's the correct solution.

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    $\begingroup$ It is stated that the $t$ is the time after the first launch. Therefore your second integral should go from $4$ to $x$. $\endgroup$ – maxmilgram Jan 9 at 9:04
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The velocity of the first rocket is $v_1(t) = 6 − t$ metres per second and the velocity of the second is $v_2(t) = 10 − t$ metres per second,where $t$ is the time in seconds after the first launch.

The time $t$ in the second velocity function is also counted as seconds after the first launch so in your integral, you should not start from $t=0$ (and end at $t=x-4$) but start at $t=4$ (and end at $t=x$), to get: $$\int_0^x v_1(t) \,\mbox{d}t = \int_\color{red}{4}^\color{red}{x} v_2(t) \,\mbox{d}t \iff \int_0^x \left(6-t\right) \,\mbox{d}t = \int_\color{red}{4}^\color{red}{x} \left(10-t\right) \,\mbox{d}t \iff x = 8$$

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  • $\begingroup$ Thanks for your quick reply. I got it :). $\endgroup$ – 王文军 or Wenjun Wang Jan 9 at 9:11
  • $\begingroup$ Alright, you're welcome! $\endgroup$ – StackTD Jan 9 at 9:11

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