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The displacement of a particle moving in a straight line is represented by ax^2 +3x =t, (where x is displacement).At an instant when v=1ms acceleration is found to be -4ms^-2 then a=?

Santosh Rao , 6 Years ago
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anser 2 Answers
bhanutej

Last Activity: 6 Years ago

ax2+3x=t
differentiate wrt t
2ax+3v=1       [v=dx/dt,A=dv/dt]
2ax=1-3v
2ax=1-3(1)=-2  -------->(!)
diff again wrt
2a(v.v+xA)+3A=0
2av2+2axA+3A=0
2a.1+(-2A)+3A=0     [from -->(1)]
2a=-A
2a=2
a=1
 
 

Ashwani kumar

Last Activity: 6 Years ago

 
t=ax^2+3x
dt/dx=2ax+3=1/v
2ax+3=1
ax=1
v=1/2ax+3
squaring both sides
v2=(2ax+3)2
2v.dv/dx=-2(2ax+3)-3 ×(2a×1+0)
a=-2a/(2ax+3)3
a=-2a
bua a=-4
-4=-2a
a=2 ans.
 

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