Question

if the velocity "v" of a body os given by =√64+5x when its displacement is x then its acceleration is​

Answer 1

We know that acceleration is rate of change of velocity.

We have, v=(180–16x)^1/2

Now,differentiating it w.r.t t i.e time,

So, dv/dt = [1/2(180–16x)^(-1/2)](-16dx/dt)

(I hope u have no problem in differentiating, if u have then please recinsider because it's basics)

So, on simplifying above equation,

a(acc.)=dv/dt=[1/2(180–16x)^(-1/2)](-16v)

Putting the value of v from original equation,

a=dv/dt=[-8(180–16x)^(-1/2)](180–16x)^1/2)

Clearly,on simplifying we get,

a=dv/dt=-8 m/sec^2 which is the required answer.
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1 comment from Gurkirat Singh

Sandeep Bangra, Physics Faculty at IMA Jodhpur (2015-present)
Answered August 12, 2019

Answer 2

Answer:

The acceleration of the body will be equal to 2.5ms⁻².

Explanation:

We have given an expression for the velocity of the body:

$v=\sqrt{64+5x}$                                                           ................(1)

where x is displacement.

We know that the acceleration is rate of change of velocity per unit time.

So, differentiate the equation (1) w.r.t. time 't',

$a=\frac{dv}{dt}$

$a=\frac{d}{dt}(\sqrt{64+5x })$

$a=\frac{1}{2}(64+5x)^{\frac{1}{2}-1}\frac{d}{dt}(64+5x)$

$a=\frac{1}{2}(64+5x)^\frac{-1}{2}(0+5\frac{dx}{dt} )$

$a=\frac{1}{2}(64+5x)^\frac{-1}{2}(5v)$

Substitute the value of 'v' in above equation:

$a=\frac{1}{2}(64+5x)^\frac{-1}{2}[5(64+5x)^\frac{1}{2}]$

By Simplify the above equation, we get;

$a=\frac{5}{2}$

a = 2.5ms⁻²

Therefore, the acceleration of the body us equal to 2.5ms⁻².