Answer :
Answer:
285 seconds
Explanation:
Jenny speed is 3.8 m/s
Alyssa speed in 4.0 m/s
Alyssa starts after 15 seconds
Find the distance covered by Jenny, when Alyssa starts
Distance=Speed*time
Distance covered by Jenny in 15 seconds= 3.8×15=57m
Relative speed of the two members heading same direction will be;
4.0m/s-3.8m/s=0.2m/s
To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two
Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds
=4.75 minutes
Answer:
After 100 seconds Alyssa catch up with Jenny.
Explanation:
Jenny's data:
[tex]v_{Jenny} =3.8m/s[/tex]
[tex]t_{Jenny}=t[/tex]
[tex]d_{Jenny}=d[/tex]
Alyssa's data:
[tex]v_{Alyssa}=4.0m/s[/tex]
[tex]t_{Alyssa}=t-15[/tex], because she has a difference of 15 seconds.
[tex]d_{Alyssa}=d[/tex]
Both move at a constant speed, that means there's no acceleration, their speed is always the same.
Now, the equation of each movement is
[tex]d=3.8t[/tex] and [tex]d=4(t-15)[/tex], then we solve this two.
We replace the first equation into the second one
[tex]3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100[/tex]
That means after 100 seconds Alyssa catch up with Jenny.