Question

a body is in straight line motion with an acceleration given by a=32 - 4v.the initial conditions are t=0 v=4 find the velocity when t=ln 2

Answer 1

Answers : (1)
dv/(32-4v) = dt..

integrate

lhs from 4 to v

rhs from 0 to ln2

thus

-4ln(32-4v)]v,4 = ln2

32-4v]v,4 = 2^(-1/4)

32-4v -32 + 16 = 0.85

4v = 15approx

v = 3.7

Answer 2

Answer:

Concept:

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector.

                           ${\displaystyle {\boldsymbol {\bar {v}}}={\frac {\Delta {\boldsymbol {x}}}{\Delta t}}.}$

Explanation:

Given:

A body is in straight-line motion with an acceleration given by a=32 - 4v

The initial conditions are t=0 v=4

Find:

The velocity when t=ln 2
Solution:

                     $\frac{\mathrm{dv}}{(32-4 \mathrm{v})}=\mathrm{dt}$

Integrate: -

L.H.S from 4 to $\mathrm{v}$

R.H.S from 0 to $In} 2$

Thus,

                       $\begin{aligned}&(32-4 v) v, 4=2 \\&32-4 v-32+16=0.85 \\&4 v=15 \\&v=\frac{15}{4}\end{aligned}$

The velocity is  $\frac{15}{4}$

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