Jabal and Michael are walking to school and agree to leave their homes at the same time. Jabal leaves his house walking 2.5 meters per second, Michael leaves his house walking 6 meters per second
Jabal's house is 15 meters closer to school than Michael's house, therefore he is starting 16 meters closer than Michael
After how long are the boys walking together? Use any strategy you'd like to solve

We can write the position equation as:

P(t) = speed*time + initial position.

If we define Jabal home as the 0 of the position, then Jabal's initial position is 0m

And we know that his speed is 2.5 m/s, then his position equation is:

Pj(t) = (2.5 m/s)*t

We know that Michael speed is 6m/s, and we also know that his house is 15m further away from school than Jabal's one, then Michael's initial position is -15m

Michael's position equation is:

Pm(t) = (6m/s)*t -15 m

The boys will be walking together when they meet, and this will happen when their positions are the same, this means that:

Pj(t) = Pm(t)

That is the equation we need to solve for t.

(2.5 m/s)*t = (6m/s)*t -15 m

15m = (6m/s)*t  -  (2.5 m/s)*t  = (3.5 m/s)*t

15m/(3.5 m/s) = t = 4.28 seconds.

The boys will walk together after 4.28 seconds.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: