Mixing
A Calculus Question from Math 1A Fall 2018 Practice 2
$begingroup$
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
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
$endgroup$
add a comment |
$begingroup$
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
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
$endgroup$
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
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
asked Jan 9 at 8:55
ShinobuIsMyWifeShinobuIsMyWife
183
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
$endgroup$
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:
$$int_0^x v_1(t) ,mbox{d}t = int_color{red}{4}^color{red}{x} v_2(t) ,mbox{d}t iff int_0^x left(6-tright) ,mbox{d}t = int_color{red}{4}^color{red}{x} left(10-tright) ,mbox{d}t iff x = 8$$
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
answered Jan 9 at 9:02
StackTDStackTD
22.6k2050
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
add a comment |
$begingroup$
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
add a comment |
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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