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Irodov 1.1Trickykinematicsrelative-motionreference-frames

Motorboat and Raft - River Flow Velocity

Problem 1.1

A motorboat going downstream overcame a raft at a point

A

;

\tau = 60 \text{ min}

later it turned back and after some time passed the raft at a distance

l = 6.0 \text{ km}

from the point

A

. Find the flow velocity assuming the duty of the engine to be constant.

Answer:

u = 3.0 \text{ km/h}

Solution Path

In this problem the key was switching to the raft's frame where the river vanishes. The boat goes out and returns in equal times (total 2 tau). The raft drifts l in that time, giving u = l/(2 tau) = 3.0 km/h.

01Problem Restatement

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A motorboat is traveling downstream on a river. It passes a raft floating in the water at some point

A

. The boat keeps going for a while, then turns around and heads back upstream. Eventually, it passes the same raft again, but by now the raft has drifted downstream.Given:
-

\tau = 60 \text{ min} = 1.0 \text{ h}

(time the boat traveled before turning back)
-

l = 6.0 \text{ km}

(distance from

A

to the point where the boat meets the raft again)
- The engine runs at constant power (so the boat's speed relative to water is constant)We need to find:

u

, the flow velocity of the river, in km/h.

Find

u

(river flow velocity in km/h)

02Physical Picture and Strategy

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Picture a river flowing to the right. The raft has no engine. It simply floats with the current at the river's speed

u

. The motorboat has its own speed

v

relative to the water.When the boat goes downstream, the river helps it along, so its ground speed is

v + u

. When it goes upstream, the river fights it, so its ground speed is

v - u

.The boat starts at

A

, zooms downstream for

\tau

minutes, reaches some point

B

, turns around, and comes back upstream until it meets the raft at point

C

, which is

l = 6.0 \text{ km}

from

A

.Strategy: This looks complicated in the ground frame because we have two unknowns (

u

and

v

) and would need to track both objects. Instead, we will switch to the raft's reference frame, where the river disappears entirely and the problem becomes trivial.

Ground speed:

v + u

(downstream),

v - u

(upstream). Raft drifts at

u

03Step 1: Switch to the Raft's FrameKEY INSIGHT

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The raft moves with the river. So from the raft's point of view, the water around it is stationary. There is no current.Using the principle of relative motion (Galilean transformation):In the raft's frame, the boat simply moves through still water at speed

v

. It goes away from the raft at speed

v

for time

\tau

, then comes back at speed

v

for some time.Since the speed is the same in both directions (no river to help or hinder), the return trip takes exactly the same time as the outward trip.

\text{Time away} = \tau, \quad \text{Time back} = \tau

\text{Total time from meeting to meeting} = 2\tau

This means: from the moment the boat passes the raft at

A

to the moment they meet again at

C

, exactly

2\tau

of time has passed.

In raft's frame: total time between meetings

= 2\tau

04Step 2: Raft's Displacement

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Now we switch back to the ground frame to find the raft's displacement.The raft drifts at the river's speed

u

for the entire duration

2\tau

. Using the definition of displacement for constant velocity:

l = u \cdot 2\tau

where

l = 6.0 \text{ km}

is the distance the raft has drifted from point

A

, and

2\tau = 2 \times 1.0 \text{ h} = 2.0 \text{ h}

.This means: however far the raft drifted in

2\tau

hours is exactly the

l

we were given.

l = u \cdot 2\tau

05Step 3: Solve for Flow Velocity

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We rearrange

l = 2u\tau

to isolate the unknown

u

u = \frac{l}{2\tau}

Substituting the given values:

u = \frac{6.0 \text{ km}}{2 \times 1.0 \text{ h}}

u = \frac{6.0 \text{ km}}{2.0 \text{ h}} = 3.0 \text{ km/h}

This means: the river flows at 3.0 km/h. Notice that the boat speed

v

cancelled out completely. The answer does not depend on how fast the boat is.

u = 3.0 \text{ km/h}

06Verification and Summary

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Verification: Plugging

u = 3.0 \text{ km/h}

back in:

\text{Raft drifts: } 3.0 \text{ km/h} \times 2.0 \text{ h} = 6.0 \text{ km} \; \checkmark

This matches the given distance

l = 6.0 \text{ km}

. Checks out.Dimensional check:

\frac{[\text{km}]}{[\text{h}]} = \text{km/h}

\checkmark

Summary: In this problem the key was switching to the raft's reference frame. In that frame, the river disappears and the boat simply goes out and comes back at the same speed, taking equal times. The total event lasts

2\tau

, during which the raft drifts

l

in the ground frame. The flow velocity is

l/(2\tau) = 3.0

km/h, independent of the boat speed.

\boxed{u = 3.0 \text{ km/h}}

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