The velocity of a particle moving in a straight line is directly proportional to 34th power of time elapsed- The ratio of its displacement and acceleration be tx-Find x-

Given, v ∝ t^3/4 or v=kt^3/4 Thus, a=dvdt=34kt^ 1/4 and x=∫vdt=4kt^7/47 Thus, xa=1621t^2 , thus x=2 ...

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Standard VII

Physics

Deceleration/Retardation

The velocity ...

Question

The velocity of a particle moving in a straight line is directly proportional to ${\frac{3}{4}}^{t h}$ power of time elapsed. The ratio of its displacement and acceleration be $t^{x}$ .Find $x .$

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Solution

Given, $v \propto t^{3 / 4}$ or $v = k t^{3 / 4}$
Thus, $a = \frac{d v}{d t} = \frac{3}{4} k t^{- 1 / 4}$
and $x = \int v d t = \frac{4 k t^{7 / 4}}{7}$
Thus, $\frac{x}{a} = \frac{16}{21} t^{2}$ , thus $x = 2$

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The velocity of a particle moving in a straight line is directly proportional to 34th power of time elapsed. The ratio of its displacement and acceleration be tx.Find x.

Given, v∝t3/4 or v=kt3/4Thus, a=dvdt=34kt−1/4and x=∫vdt=4kt7/47Thus, xa=1621t2, thus x=2

The velocity of a particle moving in a straight line is directly proportional to ${\frac{3}{4}}^{t h}$ power of time elapsed. The ratio of its displacement and acceleration be $t^{x}$ .Find $x .$

Given, $v \propto t^{3 / 4}$ or $v = k t^{3 / 4}$
Thus, $a = \frac{d v}{d t} = \frac{3}{4} k t^{- 1 / 4}$
and $x = \int v d t = \frac{4 k t^{7 / 4}}{7}$
Thus, $\frac{x}{a} = \frac{16}{21} t^{2}$ , thus $x = 2$