Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

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Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

) assuming that they start at the same point ant the same time, how much sooner does the faster arrive at a destination 780 m away?b.) how far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

Mathematics

1 answer:

Zina [12.1K] · Answer 1 · 2026-02-17T17:08:19+03:00

Answer:

a) The faster student arrives 7 minutes and 36 seconds earlier

b) They would need to walk 598.51 meters

Step-by-step explanation:

Speed = Distance / Time

Rearranged for time:

Time = Distance / Speed

First student:

Time = 780 m / 0.90 m/s = 866.66 seconds

Second student:

Time = 780 m / 1.90 m/s = 410.52 seconds

Difference = 866.66 - 410.52 = 456.14 seconds

Converting to minutes:

456.14 seconds / 60 = 7 minutes 36 seconds

Using Distance = Speed × Time

Given: Speed₁ × Time₁ = Speed₂ × Time₂

And that Time₁ = Time₂ + 5.50 minutes

5.50 minutes = 350 seconds

Substituting:

Time₂ = (Speed₁ × (Time₂ + 350)) / Speed₂

Time₂ = 0.9 × (Time₂ + 350) / 1.9

Time₂ = 0.4737 × (Time₂ + 350)

Time₂ = 0.4737 Time₂ + 165.79 seconds

Time₂ - 0.4737 Time₂ = 165.79 seconds

0.5263 Time₂ = 165.79 seconds

Time₂ = 165.79 / 0.5263 = 315.01 seconds

Distance = 1.9 × 315.01 = 598.51 meters

Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

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