A Calculus Question from Math 1A Fall 2018 Practice 2












3












$begingroup$


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.






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    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
















3












$begingroup$


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.






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    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














3












3








3


1



$begingroup$


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.






calculus







share|cite|improve this question









$endgroup$




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.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 8:55









ShinobuIsMyWifeShinobuIsMyWife

183




183








  • 1




    $begingroup$
    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














  • 1




    $begingroup$
    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








1




1




$begingroup$
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




$begingroup$
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










1 Answer
1






active

oldest

votes


















3












$begingroup$


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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick reply. I got it :).
    $endgroup$
    – ShinobuIsMyWife
    Jan 9 at 9:11










  • $begingroup$
    Alright, you're welcome!
    $endgroup$
    – StackTD
    Jan 9 at 9:11











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3












$begingroup$


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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick reply. I got it :).
    $endgroup$
    – ShinobuIsMyWife
    Jan 9 at 9:11










  • $begingroup$
    Alright, you're welcome!
    $endgroup$
    – StackTD
    Jan 9 at 9:11
















3












$begingroup$


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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick reply. I got it :).
    $endgroup$
    – ShinobuIsMyWife
    Jan 9 at 9:11










  • $begingroup$
    Alright, you're welcome!
    $endgroup$
    – StackTD
    Jan 9 at 9:11














3












3








3





$begingroup$


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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$






share|cite|improve this answer









$endgroup$




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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 9:02









StackTDStackTD

22.6k2050




22.6k2050












  • $begingroup$
    Thanks for your quick reply. I got it :).
    $endgroup$
    – ShinobuIsMyWife
    Jan 9 at 9:11










  • $begingroup$
    Alright, you're welcome!
    $endgroup$
    – StackTD
    Jan 9 at 9:11


















  • $begingroup$
    Thanks for your quick reply. I got it :).
    $endgroup$
    – ShinobuIsMyWife
    Jan 9 at 9:11










  • $begingroup$
    Alright, you're welcome!
    $endgroup$
    – StackTD
    Jan 9 at 9:11
















$begingroup$
Thanks for your quick reply. I got it :).
$endgroup$
– ShinobuIsMyWife
Jan 9 at 9:11




$begingroup$
Thanks for your quick reply. I got it :).
$endgroup$
– ShinobuIsMyWife
Jan 9 at 9:11












$begingroup$
Alright, you're welcome!
$endgroup$
– StackTD
Jan 9 at 9:11




$begingroup$
Alright, you're welcome!
$endgroup$
– StackTD
Jan 9 at 9:11


















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