Lesson Using vectors to solve problems in Mechanics: Velocity

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Using vectors to solve problems in Mechanics: Velocity


Problem 1

A rowboat moves across the river.  A rower in the boat heads a direction                                        
perpendicular to the river's banks  (Figure 1a).  The boat's speed in this
direction is 2 m/s.  The river current speed is 1 m/s.

Draw the vector of the full velocity of the boat relative to the river's bank.

You can assume that the river's banks are parallel straight lines.

Solution

The full vector of the boat velocity relative the river's banks has two components.
From one side,  the boat moves across the river with oars having the velocity
component  2  m/s  perpendicular to the river's banks due to rowing.
From the other side,  the boat moves along the river having the velocity component
1  m/s  in this direction relative to the river banks due to current.


Figure 1a.  The river, the current
and the rowboat crossing the river          


Figure 1b.  The full vector of the
boat velocity relative to the river banks

The  Figure 1b represents the full vector of the boat velocity relative to the river's banks.  It has the component  2  m/s in the direction perpendicular to the banks and the component  1  m/s in the direction along to the river.  The angle alpha of the  Figure 1b  is  tan%28alpha%29 = 1%2F2,  so  alpha = atan%281%2F2%29 = 0.4636 rad = 26.65°.
The full speed  (the magnitude of the velocity relative to the river banks)  is  sqrt%282%5E2%2B1%5E2%29 = sqrt%285%29 = 2.236 m/s.

Answer. The full vector of the boat velocity has the components  2  m/s  in the direction perpendicular to the river bank,  and  1  m/s  in the direction along the current.  This vector makes the angle  26.65°  with the straight line perpendicular to the river banks.  The magnitude of this vector  (the full speed of the boat relative to the river banks)  is  2.236  m/s.


Problem 2

A river whose width is  200 m,  has the current speed of  1 m/s.
A rowboat is crossing the river.  A rower heads due a direction perpendicular                                 
to the river's banks rowing at the speed of  2 m/s  relative to the water.
At what distance the rowboat will get the other bank counting downstream
from the point  O  opposed to the starting point  D  (Figure 2a)?

You can assume that the river's banks are parallel straight lines.

Solution

The rowboat has the speed of  2 m/s  in the direction perpendicular to the banks
due to rowing.  Hence,  it will take  %28200+m%29%2F%282%28m%2Fs%29%29 = 100 seconds  to cross the river
and to get the opposite bank.


Figure 2a.  The river, the current
and the rowboat crossing the river          


Figure 2b.  The full vector of the
boat velocity relative to the river banks

In  100 seconds  the boat will drift downstream with the current a distance of   1 m%2Fs * 100 s = 100 m.

Answer.  The rowboat will get the other bank in  100 m  downstream from the point  O  opposed to the starting point.

The  Figure 2b  (the same as the  Figure 1b)  represents the full vector of the boat velocity relative to the river's banks.  It has the component  2  m/s in the direction perpendicular to the banks and the component  1  m/s in the direction along to the river.  The angle alpha of the  Figure 2b  is  tan%28alpha%29 = 1%2F2,  so  alpha = atan%281%2F2%29 = 0.4636 rad = 26.65°.


Problem 3

A man is going to cross the river in a rowboat.  The river's width is  200 m,                                    
the current speed is of  1 m/s.
A man can row at the speed of  1 m/s  relative to the water.
At what direction should the rower head the rowboat to get the other bank
at the point A  directly opposed to the starting point  D  (Figure 3a)?

You can assume that the river's banks are parallel straight lines.

Solution

Because the boat is moving partly with the current,  the boat headed due
the perpendicular to the river's banks will drift with the current.  To
compensate for this cross-flow,  the boat must be headed due some angle
against the current as it is shown in the Figures  3a  and  3b.


    Figure 3a.  The rowboat crosses          
    the river overcoming the current


      Figure 3b.  The vectors of the
                    velocities

The velocity of the rowboat relative to the banks  Vrb  will be the sum of the velocity of the rowboat relative to the water  Vrw  and the velocity of the current relative to the river's banks  Vcb:   Vrb = Vrw + Vcb.

The magnitudes of the velocities  |Vrw|  and  |Vcb|  are  |Vrw| = 2 m/s  and  |Vcb| = 1 m/s.

From the right triangle of the  Figure 3b  we have for the angle  alpha:   sin%28alpha%29 = abs%28Vcb%29%2Fabs%28Vrw%29 = 1%2F2.  Hence,  alpha = arcsin%281%2F2%29 = 30°.

From the same right triangle of the  Figure 3b  we have for the velocity magnitude  |Vrb|:   |Vrb| = sqrt%28Vrw%5E2+-+Vcb%5E2%29 = sqrt%282%5E2+-+1%5E2%29 = sqrt%284-1%29 = sqrt%283%29 =~ 1.732 m/s.

Answer. The rower must head the rowboat due the direction that makes the angle  30°  with the perpendicular to the river's banks to get the other bank at the point directly opposed to the starting point.  Under the given conditions,  the rowboat speed perpendicular the the river's banks will be  1.732 m/s.  It will takes  %28200+m%29+%2F%281.732+%28m%2Fs%29%29 = 115.5  seconds  for the rowboat to cross the river.


Problem 4

In rainy weather the drivers who drive their cars on the highway,  can see the slanting                
raindrops behind the car's side windows  (Figure 4a).
Calculate the angle that the raindrops makes with the vertical  (Figure 4b)  as
viewed by the driver,  if the raindrop speed near the earth surface is  10 m/s  
(meters per second)  and the car's speed is  65  mph  (miles per hour).
Assume that raindrops fall vertically down  in the coordinate system connected
to the earth.

Solution

First,  let us convert the speed of  65  mph to meters per second.
1 mile = 1609 m (approximately),  so  65 mph = 1609*65/3600 = 29.05 m/s.


  Figure 4a.  The slanting raindrops
     behind the car's side window              


    Figure 4b.  The angle between
     the raindrops and the vertical

Relative to the moving car the falling raindrop has the horizontal velocity  29.05 m/s  in the direction opposite to the car move.  It has the vertical velocity of  10 m/s,  too.  So,  the velocity of the falling raindrop relative to the car is the vector with the horizontal component  29.05 m/s  and the vertical component  10 m/s.  Hence,  we have for the angle  alpha  with the vertical  (Figure 4b)  tan%28alpha%29 = 29.05%2F10 = 2.905  and  alpha = atan%282.905%29 = 1.239 rad = 71°.

Answer.  The angle that the raindrops makes with the vertical as viewed by the driver is  71°.


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