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Take-home experiment: relative velocity of a boat

Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.

Adding velocities: a boat on a river

A boat is trying to cross a river. Due to the velocity of the river the path traveled by the boat is diagonal. The velocity of the boat, v boat, is equal to zero point seven five meters per second and is in positive y direction. The velocity of the river, v-river, is equal to one point two meters per second and is in positive x direction. The resultant diagonal velocity v total, which makes an angle of theta with the horizontal x axis, is towards north east direction.
A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the boat relative to the shore?

Refer to [link] , which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, v tot size 12{v rSub { size 8{"tot"} } } {} . The velocity of the boat, v boat size 12{v rSub { size 8{"boat"} } } {} , is 0.75 m/s in the y size 12{y} {} -direction relative to the river and the velocity of the river, v river size 12{v rSub { size 8{"river"} } } {} , is 1.20 m/s to the right.

Strategy

We start by choosing a coordinate system with its x -axis parallel to the velocity of the river, as shown in [link] . Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the y -axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations v tot = v x 2 + v y 2 size 12{v rSub { size 8{"tot"} } = sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ = tan 1 ( v y / v x ) size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {} directly.

Solution

The magnitude of the total velocity is

v tot = v x 2 + v y 2 , size 12{v rSub { size 8{"tot"} } = sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } ","} {}

where

v x = v river = 1 . 20 m/s size 12{v rSub { size 8{x} } =v rSub { size 8{"river"} } =1 "." "20"" m/s"} {}

and

v y = v boat = 0 . 750 m/s. size 12{v rSub { size 8{y} } =v rSub { size 8{ ital "boat"} } =0 "." "750 m/s."} {}

Thus,

v tot = ( 1 . 20 m/s ) 2 + ( 0 . 750 m/s ) 2 size 12{v rSub { size 8{"tot"} } = sqrt { \( 1 "." "20"" m/s" \) rSup { size 8{2} } + \( 0 "." "750"" m/s" \) rSup { size 8{2} } } } {}

yielding

v tot = 1 . 42 m/s. size 12{v rSub { size 8{"tot"} } =1 "." "42"" m/s."} {}

The direction of the total velocity θ size 12{θ} {} is given by:

θ = tan 1 ( v y / v x ) = tan 1 ( 0 . 750 / 1 . 20 ) . size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) ="tan" rSup { size 8{ - 1} } \( 0 "." "750"/1 "." "20" \) "."} {}

This equation gives

θ = 32 . . size 12{θ="32" "." 0º} {}

Discussion

Both the magnitude v and the direction θ of the total velocity are consistent with [link] . Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 32.0º size 12{"32.0º"} {} ) the total velocity has relative to the riverbank.

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Calculating velocity: wind velocity causes an airplane to drift

Calculate the wind velocity for the situation shown in [link] . The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20 .0º size 12{"20" "." 0 rSup { size 8{o} } } {} west of north.

An airplane is trying to fly north with velocity v p equal to forty five meters per second at angle of one hundred and ten degrees but due to wind velocity v w in south west direction making an angle theta with the horizontal axis it reaches a position in north west direction with resultant velocity v total equal to thirty eight meters per second and the direction is twenty degrees west of north.
An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind?

Strategy

In this problem, somewhat different from the previous example, we know the total velocity v tot size 12{v rSub { size 8{ bold "tot"} } } {} and that it is the sum of two other velocities, v w size 12{v rSub { size 8{w} } } {} (the wind) and v p size 12{v rSub { size 8{p} } } {} (the plane relative to the air mass). The quantity v p size 12{v rSub { size 8{p} } } {} is known, and we are asked to find v w size 12{v rSub { size 8{w} } } {} . None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of v w size 12{v rSub { size 8{w} } } {} , then we can combine them to solve for its magnitude and direction. As shown in [link] , we choose a coordinate system with its x -axis due east and its y -axis due north (parallel to v p size 12{v rSub { size 8{p} } } {} ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods .)

Questions & Answers

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Xolani Reply
movement in a direction
Jason
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Yimam
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thomas Reply
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
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Promise
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Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
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Oluwakayode
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yes
haider
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pressure
Oluwakayode
Dimension for force MLT-2
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Osueke Reply
how do you calculate the 5% uncertainty of 4cm?
melia Reply
4cm/100×5= 0.2cm
haider
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melia Reply
= 200g±(5%)10g
haider
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melia
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haider
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haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
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melia
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Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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