# A particle is st in motion at the moment t=0, at the velocity v0 wher upon its velocity begins changing with time in accordance with the law.v=v0(1-t/b) b=constant Find : 1. The displacement dr...

A particle is st in motion at the moment t=0, at the velocity v0 wher upon its velocity begins changing with time in accordance with the law.

v=v0(1-t/b) b=constant

Find :

1. The displacement dr of the particle 2

2. The distance is covered by it in the first t seconds of motion

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### 2 Answers

Given that velocit v = v0(1-t/b), where b is aconstant.

To find the distance s(t) during the time t.

We know that velocity v = ds/dt.

Therefore ds/dt = v0 (1-t/b).

ds = v0(1-t/b)dt

ds = v0dt -t/dt

We integrate with respect to the variable time t.

Integral ds = Integral v0dt - Integral (t/b) dt

s(t) = v0t - t^2/2b + const.

WE presume at = 0, s(0) = 0.

So 0 = v0*0 -0^2/(2b) +C .

Therefore 0 = 0+C. .

So C = 0.

Therefore the distance formula is s(t) = v0*t - (1/2b) t^2 is the expression for the distance traversed in time t.

If you wish to find the particle at travelled between the any two time interval t1 and t2 we can find s(t1)- s(t1) = v0(t2-t1)- (1/2b)(t2-t1).

1) We know that the space is the product of velocity and time.

r = v*t

We'll differentiate both sides:

dr = vdt

To calulate r = Integral vdt

We'll substitute v by the given expression:

Integral vdt = Int v0(1-t/b)dt

Int v0(1-t/b)dt = v0*Int(1-t/b)dt

v0*Int(1-t/b)dt = v0*(Int dt -Int tdt/b)

v0*(Int dt -Int tdt/b) = v0*t - (v0/b)*t^2/2 + C

**r = v0*t - (v0/b)*t^2/2 + C**

2) We'll calculate the distance covered by the particle in time:

s = Int vdt

In this case v = v0*|1-t/b|

If t<b, we'll have v = v0*(1 - t/b)

If t>b, we'll have v = v0*(t/b - 1)

If **t<b**

s = Int v0*(1 - t/b) dt

**s = v0*t - (v0/b)*t^2/2 + C**

If** t>b**

s = Int v0*(t/b - 1) dt

**s = v0*b{1+[1-(t/b)]^2}/2 + C**