A train is moving with a velocity of 30 km/hr due east and a car is moving with a velocity of 40 km/hr due north. What is the velocity of car as  appears to a passenger in the train? 

Velocity of train, vT = 30 km /hr

Velocity of car, vc = 40 km/hr

To find: relative velocity of car w.r.t. train. To bring the train at rest, apply equal and opposite velocity of train on car, which is 30 km/hr. 

So, relative velocity of car w.r.t train is, 

vCT = OD = (OB2 + BD2)1/2

               = (402 + 302)1/2 

               = 50 km/hr 

Let <DOB = straight theta, then 

 tan space straight theta space equals space BD over OB space equals space 30 over 40 space equals space 3 over 4 space
space space space space space space space space space space equals space 0.75 space equals space tan space 36 to the power of straight o space 52 apostrophe

space space space space space space space space space space equals space 36 to the power of straight o space 52 apostrophe comma space west space of space north. space 

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Two trains 120 m and 80 m in length are running in opposite directions with velocities 42 km/hr and 30 km/hr. In what time they will completely cross each other?

Relative velocity of one train w.r.to second is, 

= 42 - (-30)

= 72 km/hr = 20 m/s

Total distance to be travelled = 120 +80 = 200 m

Time taken, t = 200 over 20 space equals space 10 space s 

In 10 sec, the two trains will completely cross each other. 

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Two persons A and B are walking with speed 4 km/hr and 5 km/hr respectively in the same direction. Find how far will be B from A after 3 hours? 


Relative velocity of B w.r.to A,  

vBA = vB - vA

      = 5- 4 = 1 km/hr 

Therefore, distance of B ahead of A in time is, 

d = vAB x t

   = 1 x 3

    = 3 km

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The blades of an aeroplane propeller are rotating at the rate of 600 revolutions per minute. Calculate its angular velocity? 

Here, 

Angular frequency, straight nu = 600 r.pm 

                                = 600 over 60 space r e v o l u t i o n divided by s e c 

We know, 

straight omega space equals space 2 πν space equals space 2 straight pi space straight x space 600 divided by 60 space

space space space space equals space 20 space straight pi space rad divided by sec

This is the required angular velocity. 

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A particle starts with a velocity of 200 cm/s and moves in a straight line with a retardation of 10 cm/s2. Find the time it takes to describe 1500 cm and explain the double answers. 

Here, we have

Initial velocity, u = 200 cm/s 

Acceleration, a = -10 cm/s2 

Distance travelled, s = 1500 cm 

Time taken, t = ? 

Using the relation, 

        s = ut + 1/2 at2

  1500 = 200 t + 1/2(-10)t2

On solving the equation,we get

t = 10 s or t = 30 s

Here, the value of 10 s corresponds to the time when the particle first arrives at the given location. It then crosses this point and at the end of 20 s, its velocity is just 0. 

The particle then returns and at the end of 10 s more, is again at the given location from the starting point. 

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