the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is?
1sec 2 sec 3 sec or 4 sec

Question

the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is?

1sec 2 sec 3 sec or 4 sec

Vikas TU · Answer

Given, a=-v(ln(2^(1/2)) a = – (2.303v/2)*log2 = > -0.3v dv/v = – 0.3dt inetegerating both sides we get, logv – log(v/2) = – 0.3t -log2 = -0.3t t = 1 sec.

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the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is? 1sec 2 sec 3 sec or 4 sec

the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is?
1sec 2 sec 3 sec or 4 sec

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the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is? 1sec 2 sec 3 sec or 4 sec

the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is?1sec 2 sec 3 sec or 4 sec

1 Answers

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the accelaration of a particle moving in a straight line depends on its velocity as a=-v(ln(2^(1/2)). the time interval where velocity becomes half of its value at beginning is?
1sec 2 sec 3 sec or 4 sec