Answered

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Ricardo and John start swimming from the same location. Ricardo starts 15 seconds before John and
swims at a rate of 3 feet per second. John swims at a rate of 4 feet per second in the same direction as
Ricardo. Which equation could you solve to find how long it will take John to catch up with Ricardo?
Does anyone know the answer?

Ricardo And John Start Swimming From The Same Location Ricardo Starts 15 Seconds Before John And Swims At A Rate Of 3 Feet Per Second John Swims At A Rate Of 4 class=

Sagot :

Answer:

t = (t₁ · vr)/ (vj - vr) = 45 seconds

Step-by-step explanation:

Given:

Initial time from Ricardo  t₁ = 15 seconds, Ricardo's velocity (speed) vr = 3 ft/s,

John's velocity (speed) vj = 4 ft/s

Requested time when John catch Ricardo t = ?

During 15 seconds, Ricardo moved away

S = vr · t₁ = 3 ft/s · 15 s = 45 feet.

From that moment, the same time runs for both of them

Ricardo's path is S₁ = vr · t  and John's path is S₂ = vj · t

S + S₁ = S₂ =>  S = S₂ - S₁  => 45 = vj · t - vr · t  

When we draw t we get:

t ( vj - vr) = 45   t = 45 / (vj - vr) = 45 / (4 - 3) = 45 / 1 = 45 seconds

t = 45 seconds

S₁ = vr · t = 3 · 45 = 135 ft

S₂ = vj · t = 4 · 45 = 180 ft

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