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Q. A particle starts from rest with constant acceleration 'a' and it is then decelerated with a constant value 'b' till it is brought to rest.if total time taken between two rest positions is t.what is maximum speed acquired by the particle?
Ans)ab(t)/a+b

jiten , 7 Years ago
Grade 11
anser 1 Answers
Lakshya

Last Activity: 7 Years ago

The motion can be divided into two parts 
1. When accelaration is equal to a.
2. When acceleration is equal to b.
During first motion let v be its final speed and x be its time taken 
Then
V=u+at
V=0+ax
X=v÷a      (eq 1)
And in second motion v is its initial speed and final speed is 0 and let y be time taken.
 
 
Therefore
V=u+at
0=v-by because acc is negative
 
Y=v÷b    (eq 2)
By adding eq 1 and eq 2
X+y= v(a+b)÷ab
T=v(a+b)÷ab    because total time is given as T
V=abT÷(a+b)
 
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