The initial velocity of a particle is u(t = 0) and the acceleration is given by $a t^{3 / 2}$ . Which of the following relations is valid

v = u + a t^{3 / 2}

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v = u + \frac{3 a t^{3}}{2}

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v = u + \frac{2}{5} a t^{5 / 2}

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v = u + a t^{5 / 2}

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Solution

The correct option is C $v = u + \frac{2}{5} a t^{5 / 2}$
$A c c e l e r a t i o n = \frac{d v}{d t}$
$d v = A c c e l e r a t i o n d t$
$\int_{u}^{v} d v = \int_{0}^{t} a t^{3 / 2} d t$
$v - u = \frac{2}{5} a t^{5 / 2}$
$v = u + \frac{2}{5} a t^{5 / 2}$

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Q. the initial velocity of particle is u (a=0) and the accelaration is given by bt^3/2. which of the following relationship is valide????
1) v= u+bt^3/2
2) v= u+3bt³/2
3) v= u+2/5bt^5/2
4) v= u+bt^5/2
the answer is 3th option.....
how????????

Q. The initial velocity of a particle is

u

(at

t = 0

) and the acceleration a is given by

α t^{3 / 2}

. Which of the following relations is valid?

Q. Check the accuracy of relation

v^{2} - u^{2} = 2 a s

, where

v

and

u

are final and initial velocities,

a

is the acceleration, and s is the distance.

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The initial velocity of a particle is u(t = 0) and the acceleration is given by at3/2 . Which of the following relations is valid

The correct option is C v = u + 25at5/2 Acceleration=dvdtdv=Accelerationdt∫vudv=∫t0at3/2dtv−u=25at5/2v=u+25at5/2

The initial velocity of a particle is u(t = 0) and the acceleration is given by $a t^{3 / 2}$ . Which of the following relations is valid

The correct option is C $v = u + \frac{2}{5} a t^{5 / 2}$
$A c c e l e r a t i o n = \frac{d v}{d t}$
$d v = A c c e l e r a t i o n d t$
$\int_{u}^{v} d v = \int_{0}^{t} a t^{3 / 2} d t$
$v - u = \frac{2}{5} a t^{5 / 2}$
$v = u + \frac{2}{5} a t^{5 / 2}$