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Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?


In case of train A: 

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 s

Since the train is moving with uniform velocity, 

Acceleration, aI = 0

Now, using second equation of motion, 

Distance covered by train A is given by, 

straight s space equals space ut space plus space left parenthesis 1 divided by 2 right parenthesis straight a subscript 1 straight t squared space

space space space equals space 20 cross times 50 plus space 0 space equals space 1000 space straight m 

For train B, 

Initial velocity, uB72 km/h = 20 m/s

Acceleration, a = 1 m/s

Time, t = 50 s

Now, using second equation of motion, 

Distance travelled by second train B is, 

straight s subscript II space equals space ut space plus space 1 half at squared space
space space space space space equals space 20 space straight X space 50 space plus space left parenthesis 1 divided by 2 right parenthesis space cross times space 1 space cross times space left parenthesis 50 right parenthesis squared space

space space space space space equals space 2250 space straight m space

Length of both trains = 2 cross times 400 space straight m space equals space 800 space straight m
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.




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On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?

Velocity space of space car space straight A comma space straight v subscript straight A space end subscript equals space 36 space km divided by straight h space equals space 10 space straight m divided by straight s space

Velocity space of space car space straight B comma space straight v subscript straight B space equals space 54 space km divided by straight h space equals space 15 space straight m divided by straight s space

Velocity space of space car space straight C comma space straight v subscript straight C space equals space 54 space km divided by straight h space equals space 15 space straight m divided by straight s

Relative space velocity space of space car space straight B space with space respect space to space car space straight A comma space space

straight v subscript BA space equals space straight v subscript straight B space – space straight v subscript straight A space equals space 15 space – space 10 space equals space 5 space straight m divided by straight s space

Relative space velocity space of space car space straight C space with space respect space to space car space straight A comma space space

straight v subscript CA space end subscript equals space straight v subscript straight C space – space left parenthesis – space straight v subscript straight A right parenthesis space equals space 15 space plus space 10 space equals space 25 space straight m divided by straight s space

At space straight a space certain space instance comma space both space cars space straight B space and space straight C space are space at
space the space same space distance space from space car space straight A space straight i. straight e. comma space space straight s space equals space 1 space km space equals space 1000 space straight m space

space Time space taken space left parenthesis straight t right parenthesis space by space car space straight C space to space cover space 1000 space straight m space equals space 1000 space divided by space 25 space equals space 40 space straight s space

Hence comma space to space avoid space an space accident comma space car space straight B space must space cover
space the space same space distance space in space straight a space maximum space of space 40 space straight s.

space From space second space equation space of space motion comma

minimum space acceleration space left parenthesis straight a right parenthesis space produced space by space car space straight B space can
space be space obtained space as colon

space straight s space equals space ut space plus space left parenthesis 1 divided by 2 right parenthesis at squared space 1000 space

space space space equals space 5 space cross times space 40 space plus space left parenthesis 1 divided by 2 right parenthesis space cross times space straight a space cross times space left parenthesis 40 right parenthesis squared space straight a

space space space equals space 1600 space divided by space 1600 space equals space 1 space ms to the power of negative 2 end exponent
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A player throws a ball upwards with an initial speed of 29.4 m s-1.

To what height does the ball rise and after how long does the ball return to the player’s hands ?
(Take g = 9.8 m s-2and neglect air resistance).

Initial velocity of the ball, u = 29.4 m/s 

At maximum height, the velocity of the ball becomes zero.

Final velocity of the ball, v = 0 


Acceleration, a = – g = – 9.8 m/s2  

Using third equation of motion, 

Height to which the ball rises is given by, 

straight v squared space minus space straight u squared space equals space 2 gs

straight s space equals space fraction numerator left parenthesis straight v squared minus straight u squared right parenthesis over denominator 2 straight g end fraction space

space space equals fraction numerator space left square bracket left parenthesis 0 right parenthesis squared space minus space left parenthesis 29.4 right parenthesis squared right square bracket space over denominator 2 cross times left parenthesis negative 9.8 right parenthesis end fraction space space equals 44.1 space straight m
Also comma space

Time space of space ascent space equals space Time space of space desent space

So comma space total space time space taken space by space the space ball space to space return space to space the space
p layer apostrophe straight s space hand space is comma

straight t space equals space 3 space plus space 3 space equals space 6 space sec


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Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

Let V be the speed of the bus running between towns A and B.

Given, 

Speed of the cyclist  = 20 km/hr 

Relative speed of bus moving in the direction of the cyclist = V - v = (V-20) km/hr

Every 18 mins, the bus went past the cyclist. moving in the direction of the bus. 

That is, 18 over 60 space h r s
Distance covered by the bus = left parenthesis straight V minus 20 right parenthesis cross times 18 over 60 space km space space space space space space space... space left parenthesis straight i right parenthesis space
One bus leaves every t minutes. 

Therefore, 

Distance travelled by the bus = V cross times space straight T over 60     ... (ii) 

Equations (i) and (ii) are equal. 

left parenthesis straight V space minus space 20 right parenthesis space cross times space fraction numerator 18 space over denominator space 60 end fraction space equals space fraction numerator VT space over denominator 60 end fraction space space space space space space space space... left parenthesis iii right parenthesis

Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h

Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h 

Therefore, 

left parenthesis straight V plus 20 right parenthesis space cross times space 6 over 60 space equals space VT over 60 space space space space space space... space left parenthesis iv right parenthesis

Now, from equations (3) and (4), we get

space space space space space space space space space space left parenthesis straight V space plus space 20 right parenthesis space cross times space 6 space divided by space 60 space equals space left parenthesis straight V space minus space 20 right parenthesis space cross times space 18 space divided by space 60

space rightwards double arrow space space space space space space space space space straight V space plus space 20 space equals space 3 straight V space minus space 60 space

rightwards double arrow space space space space 2 straight V space equals space 80 space straight V space equals space 40 space km divided by straight h
Putting space the space value space of space straight V space in space equation space left parenthesis iv right parenthesis comma space we space get space

left parenthesis 40 space plus space 20 right parenthesis space cross times space 6 space divided by space 60 space equals space 40 straight T space divided by space 60

space space space space space space space space space space space straight T space equals space 360 space divided by space 40 space equals space 9 space min


309 Views

A player throws a ball upwards with an initial speed of 29.4 m s-1

Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive.

During downward motion, the signs of position, velocity, and acceleration are all positive.
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