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A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop.


Initial space velocity space of space the space car comma space straight u equals 126 space km divided by hr space equals space 35 space straight m divided by straight s

Final space velocity space of space car comma space straight v space equals space 0 space

Before space coming space to space rest comma space

Distance space covered space by space the space car comma space straight s space equals space 200 space straight m space

Let comma space retardation space produced space in space the space car space equals space straight a

Now comma space using space the space third space equation space of space motion comma space

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

rightwards double arrow space left parenthesis 0 right parenthesis squared space minus space left parenthesis 35 right parenthesis squared space equals space 2 space cross times space straight a space cross times space 200

rightwards double arrow space straight a space equals space minus 35 cross times 35 over 2 cross times 200 space equals space minus 3.06 space straight m divided by straight s squared space

Time space taken space by space the space car space to space stop space is space given space by space the space first space equation space of space motion. space

That space is comma space

space space space space space space space straight v space equals space straight u space plus space at

rightwards double arrow space space space straight t space equals space fraction numerator left parenthesis straight v minus straight u right parenthesis over denominator straight a end fraction space equals space fraction numerator left parenthesis negative 35 right parenthesis over denominator left parenthesis negative 3.06 right parenthesis end fraction space equals space 11.44 space sec
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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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A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

Speed space of space jet space airplane comma space straight v subscript jet space equals space 500 space km divided by hr space

relative space spedd space of space product space of space combustion
space straight w. straight r. to space the space plane space is comma space

straight v subscript smoke space equals space minus 1500 space km divided by hr space

Speed space of space the space product space of space combustion space straight w. straight r. to space ground space equals space straight v apostrophe subscript smoke

Relative space speed space of space space products space of space combustion space
with space respect space to space the space airplane comma

space space space space space straight v subscript smoke space equals space straight v prime subscript smoke space – space straight v subscript jet space end subscript

space space space – space 1500 space equals space straight v prime subscript smoke space end subscript – space 500 space straight v prime subscript smoke space end subscript

space space space space space space space space space space space space space space space space space equals space – space 1000 space km divided by straight h space

Negative sign implies that the direction of its products of combustion is opposite to the direction of motion of the jet airplane. 
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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 drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.


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