A ship P is sailing due east at 30 km/h and another ship Q is going due south with a velocity of 40 km/h. Find the velocity of ship Q that will appear to an observer in P.

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#6 {main}</pre> be the unit vectors along east and south.

Let  be the unit vectors along east and south.The velocity of ship
The velocity of ship P,       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>
The velocity of ship Q,      stack straight v subscript straight Q with rightwards arrow on top space equals space 40 straight j with hat on top
                 straight V with rightwards arrow on top space equals space stack straight v subscript straight Q with rightwards arrow on top minus stack straight v subscript straight P with rightwards arrow on top space equals space 40 straight j with hat on top minus 30 straight i with hat on top
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#6 {main}</pre>
and       space space space tanθ equals 30 over 40 equals 0.75                  rightwards double arrow space space straight theta asymptotically equal to 37 degree
Thus the velocity of ship Q w.r.t. P is 50 km/hr due 37° west of south.

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Two trains are moving in opposite directions on parallel tracks, one with a velocity twice that of the other. If the total length of two trains is 480 ft and they take 3 seconds to cross each other, find their individual velocities. 


Let the two trains A and B be moving in the opposite directions with velocities u and 2u respectively.

Relative velocity, V = u + 2u = 3u 

Total distance moved = total length of two trains.

i.e. S = 480 ft.

Time taken by trains to cross each other, t = 3 sec

Now, S     = Vt

rightwards double arrow    480 = 3u x 3

rightwards double arrow      2u = 106.66 ft/s 

rightwards double arrow        u = 53.33ft/s 



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A jet airplane travelling at the speed of 500 km/hr ejects its products of combustion at the speed of 1500 km/hr relative to the jet plane. What is the speed of the later with respect to an observer on the ground?

Let the velocity of ejecting gases with respect to ground be straight V subscript straight g . 

Velocity of jet plane,    straight V subscript straight j equals plus 500 space km divided by hr

Velocity of ejected products w.r.t. jet plane, 

               straight V subscript gj equals negative 1500 space km divided by hr.

Now, relative velocity of ejecting gases w.r.t the jet plane is given by, 

                           straight V subscript gj equals straight V subscript straight g minus straight V subscript straight j

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#6 {main}</pre>

                                equals negative 1500 plus 500 space

equals space minus 1000 space km divided by hr 

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Two ships A and B start together from the same point. A sails 16 km/h towards east and B sails 20 km/h towards north. Find the relative velocity of B w.r.t. A and distance between them after 5 hours.

Velocity of ship A,space space space stack straight v subscript straight A space with rightwards arrow on top space equals space 16 straight i with hat on top
Velocity of ship B,      stack straight v subscript straight B with rightwards arrow on top space equals space 20 straight j with hat on top

The relative velocity of B w.r.t. A is,

          straight V with rightwards arrow on top space equals space stack straight v subscript straight B with rightwards arrow on top space minus space stack straight v subscript straight A with rightwards arrow on top space equals space 20 straight j with hat on top space minus 16 straight i with hat on top

Magnitude of velocity is given by, 

∴      open vertical bar straight V with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 20 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root space equals space 25.6 space km divided by hr

and   tanθ space equals space 20 over 16 equals 1.25 space rightwards double arrow space straight theta space equals space 51.34 degree

So, relative velocity of B w.r.t. A is 25.6 km/h due <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 

The distance between two ships after 5 hours, 

          S = Vt

             = 25.6 x 5

             = 128km.

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Two trains are heading towards each other on the same straight track with speed 15 m/sec and 10 m/sec. When they are 250 m apart the drivers of both the trains apply their brakes. If one train retards at the rate 0·5 m/s2 then what should be the retardation of the second train so that collision of two trains just averts?

Initial relative speed of two trains,
straight u subscript straight r equals 15 plus 10 space equals space 25 space straight m divided by sec

Initial relative speed of two trains,Let a be the retardation of seco

Let a be the retardation of second train. Therefore the relative retardation is,
space space straight A equals left parenthesis straight a plus 0.5 right parenthesis space straight m divided by straight s squared

The collision will just avert if the relative speed of the two trains just becomes zero after travelling the distance of 250m.
Now,       space space space space straight v subscript straight r superscript 2 minus straight u subscript straight r superscript 2 space equals space 2 AS
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#6 {main}</pre>
or                   625 space equals space 500 left parenthesis straight a plus 0.5 right parenthesis
or                     straight r equals 0.75 space straight m divided by straight s squared
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