Respuesta :
Answer: 6 minutes
Explanation:
Let
[tex]t = \text{time it takes for Jill to catch up with Jack} \newline \indent d_1 = \text{distance traveled by Jill when she catches up with Jack} \newline \indent d_2 = \text{distance traveled by Jack when Jill catches him}[/tex]
Note that, Jill only catches Jack when she is faster (given in the problem) and has traveled the same distance with Jack. Mathematically,
[tex]d_1 = d_2[/tex] (1)
Moreover, since [tex]d_1[/tex] represents the distance elapsed by Jill when she catches Jack and it takes time t for Jill to do that,
[tex]d_1 = 360t[/tex] (2)
The speed of Jack is already known so we need to figure out the time he already traveled from school when Jill catches up on him to know the value of [tex]d_2[/tex].
Since Jack already walked for 2 minutes before Jill walked and it will take time t for Jill to catch up on him, Jack has already walked for time (t + 2) when he is caught up by Jill. So,
[tex]d_2 = 240(t + 2)[/tex] (3)
Then, we substitute the values of [tex]d_1[/tex] and [tex]d_2[/tex] derived in equations (2) and (3) and use equation (1) so that
[tex]360t = 240(t + 2) \\ \indent 360t = 240t + 480 \\ \indent 360t - 240t = 480 \\ \indent 120t = 480 \\ \indent t = 4 [/tex]
Hence, the time spent by Jack after he leaves school is given by
t + 2 = 6 minutes
Explanation:
Let
[tex]t = \text{time it takes for Jill to catch up with Jack} \newline \indent d_1 = \text{distance traveled by Jill when she catches up with Jack} \newline \indent d_2 = \text{distance traveled by Jack when Jill catches him}[/tex]
Note that, Jill only catches Jack when she is faster (given in the problem) and has traveled the same distance with Jack. Mathematically,
[tex]d_1 = d_2[/tex] (1)
Moreover, since [tex]d_1[/tex] represents the distance elapsed by Jill when she catches Jack and it takes time t for Jill to do that,
[tex]d_1 = 360t[/tex] (2)
The speed of Jack is already known so we need to figure out the time he already traveled from school when Jill catches up on him to know the value of [tex]d_2[/tex].
Since Jack already walked for 2 minutes before Jill walked and it will take time t for Jill to catch up on him, Jack has already walked for time (t + 2) when he is caught up by Jill. So,
[tex]d_2 = 240(t + 2)[/tex] (3)
Then, we substitute the values of [tex]d_1[/tex] and [tex]d_2[/tex] derived in equations (2) and (3) and use equation (1) so that
[tex]360t = 240(t + 2) \\ \indent 360t = 240t + 480 \\ \indent 360t - 240t = 480 \\ \indent 120t = 480 \\ \indent t = 4 [/tex]
Hence, the time spent by Jack after he leaves school is given by
t + 2 = 6 minutes
After 6 seconds if the Jack walked from school to his house at a rate of 240 ft/min. His sister Jill left school 2 minutes after Jack, walking home at a rate of 360 ft/min.
What is distance?
A numerical representation of the distance between two items or locations called distance. A physical length or an approximation based on other considerations in physics or common usage can be referred to as distance.
We have:
Jack walked from school to his house at a rate of 240 ft/min. His sister Jill left school 2 minutes after Jack, walking home at a rate of 360 ft/min.
Let's suppose the time t takes for Jill to catch up with Jack
Distance d(Ji) traveled by the Jill when she catches up with Jack
Distance d(Ja) traveled by the Jack when she catches up with Jill
According to the problem;
d(Ji) = d(Ja) ...(1)
And d(Ji) = 360t ...(2)
Speed of Jack = 360 ft/min
Jack has already walked for time (t + 2) when he is caught up by Jill. So,
d(Ja) = 240(t+2)
Now substitute the value of d(Ja) in the above equation:
360 = 240(t+2)
t = 4 seconds
and t+4 = 6 seconds
Thus, after 6 seconds if the Jack walked from school to his house at a rate of 240 ft/min. His sister Jill left school 2 minutes after Jack, walking home at a rate of 360 ft/min.
Learn more about the distance here:
brainly.com/question/26711747
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