A particle is moving in a straight line with initial velocity U and uniform accerleration A. if the sum of distances travelled in Tth & (T+1)th second is 100 cm. Then find its velocity after T second in cm/s 

Asked by SIDDHARTH SHARMA | 24th Nov, 2014, 11:08: AM

Expert Answer:

We know that the distance travelled in nth second of uniformly accelerated motion is:
begin mathsize 14px style straight S subscript straight n space equals straight u plus straight a over 2 open parentheses 2 straight n minus 1 close parentheses end style
where u is the initial velocity and a the uniform acceleration of the body.
In this case given that the sum of distances travelled in T th & (T+1) th second is 100 cm
i.e.
T +S T+1 =100
If the particle is moving with initial velocity U and uniform accerleration A,
Distance travelled by the particle in Tth second can be written as:
begin mathsize 14px style straight S subscript straight T space equals straight U plus straight A over 2 open parentheses 2 straight T minus 1 close parentheses space space space minus minus minus minus minus minus minus minus minus left parenthesis 1 right parenthesis Distance space travelled space by space the space particle space in space open parentheses straight T plus 1 close parentheses to the power of th space second space equals straight S subscript straight T plus 1 end subscript space equals straight U plus straight A over 2 open parentheses 2 open parentheses straight T plus 1 close parentheses minus 1 close parentheses straight S subscript straight T plus 1 end subscript space equals straight U plus straight A over 2 open parentheses 2 straight T plus 1 close parentheses space minus minus minus minus minus minus minus minus left parenthesis 2 right parenthesis Given space that space straight S subscript straight T space plus straight S subscript straight T plus 1 end subscript space equals 100 Substituting space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space we space get : straight U plus straight A over 2 open parentheses 2 straight T minus 1 close parentheses space plus straight U plus straight A over 2 open parentheses 2 straight T plus 1 close parentheses space space equals 100 On space solving space we space get : straight U plus AT minus straight A over 2 plus straight U plus AT plus straight A over 2 equals 100 2 straight U plus 2 AT space equals 100 rightwards double arrow straight U plus AT space equals space 50 end style
 
According to the equation of motion, we know that v = u+at
where v is the final velocity, u is the initial velocity, a the acceleration and t the time.
So comparing the equation U+AT =50 with the the equation of motion v= u+at 
we get that the final velocity of the particle at T second  = 50 cm/s
 
 

Answered by Jyothi Nair | 25th Nov, 2014, 09:27: AM

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