Question

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alyssa starts 15 s later and runs at a constant 4.0 m/s. At what time after Jenny's start does Alyssa catch up with Jenny?

Answer 1

Answer:

After 100 seconds Alyssa catch up with Jenny.

Explanation:

Jenny's data:

$v_{Jenny} =3.8m/s$

$t_{Jenny}=t$

$d_{Jenny}=d$

Alyssa's data:

$v_{Alyssa}=4.0m/s$

$t_{Alyssa}=t-15$, because she has a difference of 15 seconds.

$d_{Alyssa}=d$

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

$d=3.8t$ and $d=4(t-15)$, then we solve this two.

We replace the first equation into the second one

$3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100$

That means after 100 seconds Alyssa catch up with Jenny.

Answer 2

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