Math  /  Algebra

Question368. Mootorpaat sõitis jõel vastuvoolu 16 km ja pöördus siis tagasi, kulutades tagasiteel 40 minutit vähem aega kui liikumisel vastuvoolu. Leidke paadi kiirus seisvas vees, kui jõe voolu kiirus on 2 km/h2 \mathrm{~km} / \mathrm{h}.

Studdy Solution

STEP 1

1. The motorboat travels 16 km upstream and then returns the same distance downstream.
2. The return trip takes 40 minutes less than the upstream trip.
3. The speed of the river current is 2km/h2 \, \mathrm{km/h}.
4. We need to find the speed of the boat in still water.

STEP 2

1. Define variables for the speeds and times.
2. Write equations for the upstream and downstream trips.
3. Use the given time difference to form an equation.
4. Solve the equations to find the speed of the boat in still water.

STEP 3

Define the variables: - Let vb v_b be the speed of the boat in still water in km/h \mathrm{km/h} . - The speed of the river current is vc=2km/h v_c = 2 \, \mathrm{km/h} .

STEP 4

Write equations for the upstream and downstream trips: - Upstream speed: vbvc v_b - v_c - Downstream speed: vb+vc v_b + v_c
The time taken to travel upstream 16 km is: tup=16vb2 t_{\text{up}} = \frac{16}{v_b - 2}
The time taken to travel downstream 16 km is: tdown=16vb+2 t_{\text{down}} = \frac{16}{v_b + 2}

STEP 5

Use the given time difference to form an equation: The downstream trip takes 40 minutes (or 23\frac{2}{3} hours) less than the upstream trip, so: tuptdown=23 t_{\text{up}} - t_{\text{down}} = \frac{2}{3}
Substitute the expressions for tup t_{\text{up}} and tdown t_{\text{down}} : 16vb216vb+2=23 \frac{16}{v_b - 2} - \frac{16}{v_b + 2} = \frac{2}{3}

STEP 6

Solve the equation: Multiply through by the common denominator (vb2)(vb+2)(v_b - 2)(v_b + 2) to clear the fractions: 16(vb+2)16(vb2)=23(vb2)(vb+2) 16(v_b + 2) - 16(v_b - 2) = \frac{2}{3} (v_b - 2)(v_b + 2)
Simplify the left side: 16vb+3216vb+32=23(vb24) 16v_b + 32 - 16v_b + 32 = \frac{2}{3} (v_b^2 - 4)
64=23(vb24) 64 = \frac{2}{3} (v_b^2 - 4)
Multiply both sides by 3 to eliminate the fraction: 192=2(vb24) 192 = 2(v_b^2 - 4)
Divide by 2: 96=vb24 96 = v_b^2 - 4
Add 4 to both sides: 100=vb2 100 = v_b^2
Take the square root of both sides: vb=100 v_b = \sqrt{100} vb=10 v_b = 10
The speed of the boat in still water is: 10km/h \boxed{10 \, \mathrm{km/h}}

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