Find the Optimal Starting Time for a Tied Race in Physics Kinematics

[leprofece]

John runs at 5 m/s joe runs at 3m/seg the distance to run is 300 m Antony in order to do the race more exciting allows john start t seconds after joe What is the t value so that the race finishes in a tie?? what distance was allowed to joe?? Answer 40 seg and 120 m

I don't know how to denote the time of john
5t = 3 (300-t) ?'

 

[MarkFL]

Let's let $T$ represent the time John runs, and so $T+t$ is the time Joe runs. We know the must both run 300 m during their respective times, and so we may write:

\(\displaystyle 5T=3(T+t)\)

\(\displaystyle 5T=300\implies T=60\)

Can you proceed?

 

[leprofece]

sure 5T -3T = t
2 T = t
t = 2 *60 = 120
please I am not Sure cooecte me if I am not correct
and thanks for you reply greetings and regards

 

[MarkFL]

You should have:

\(\displaystyle 5T-3T=3t\)

You didn't distribute the $3$ before...:D

 

[leprofece]

2T = 3t

2(60) = 3t
so t = 40 seg

Ok and now: 3 (40) = 120 m

ithink i am ruight now do I?

 

[MarkFL]

2T = 3t

2(60) = 3t
so t = 40 seg

Ok and now: 3 (40) = 120 m

ithink i am ruight now do I?

Yes, that's correct! (Sun)