Question

. A body is in straight line motion with acceleration given by a = 32-4r. The intial conditions at t=0,v=4. find the velocity when t= ln2
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Answer 1

Answer:

Answer is 15/4.

Step-by-step explanation:

dv/(32 - 4v) = dt

Integrate :-

L.H.S from 4 to v

R.H.S from 0 to In2  

thus,

−4In(32−4v),4=In2

(32−4v)v,4=2  

4

−1

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32−4v−32+16=0.85

4v=15

v = 15/4

I hope that my solution have helped you.

Answer 2

Therefore the velocity of the body at time t = ln 2 is 8 units/second.

Given:

The acceleration of the body = a = 32 - 4v

Initial Conditions: t = 0; v = 4

To Find:

The velocity when t = ln 2

Solution:

The given numerical can be solved easily using the approach shown below.

Given that:  a = dv/dt = 32 - 4v

⇒ dv/dt = 32 - 4v

⇒ dv / ( 32 - 4v ) = dt

⇒ ∫ dv / ( 32 - 4v ) = ∫ dt

⇒ - ln ( 32 - 4v ) / 4 = t + c

Applying the initial conditions  t = 0; v = 4 in the above equation to find constant 'c'.

⇒ - ln ( 32 - 4 × 4 ) /4 = 0 + c

⇒ - ln ( 32 - 16 ) / 4 = c

⇒ - ln 16 / 4 = c

⇒ c = - 0.693

Substitute c = -0.693 in the above equation.

⇒ - ln ( 32 - 4v ) / 4 = t - 0.693

⇒ ln ( 32 - 4v ) = 2.77 - 4t     { Required equation for the straight equation of the body }

Now subtituting t = ln 2 in the above equation.

⇒ ln ( 32 - 4v ) = 2.77 - 4t

⇒ ln ( 32 - 4v ) = 2.77 - ( 4 * ln 2 )

⇒ ln ( 32 - 4v ) = 2.77 - 2.77

⇒ ln ( 32 - 4v ) = 0

⇒ 32 - 4v = e⁰ = 1

⇒ 4v = 32

⇒ v = 32 / 4 = 8

Therefore the velocity of the body at time t = ln 2 is 8 units/second.

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