Non-Uniform Motion With Non-Uniform Acceleration (JEE LEVEL)

Introduction

We know when the velocity of a body changes at a constant rate its acceleration is said to be constant. Similarly when the velocity of a body  changes at a non uniform rate, that means the velocity changes by different amount in same time interval the acceleration of the body is said to be non uniform or variable. It happens when the acceleration changes with either Time or Position .

Acceleration  given as a Function of Time a = f(t)

Here,               a = \frac{{dv}}{{dt}}=f\left( t \right)

\int\limits_{u}^{v}{{dv}}=\int\limits_{0}^{t}{{f\left( t \right)}}dt

or               v = u + \int\limits_{0}^{t}{{f\left( t \right)dt}}

This gives v as a function of time t i.e. we get

                  v = g(t), then

                  v = \frac{{dx}}{{dt}}=g\left( t \right)

\int\limits_{{{{x}_{o}}}}^{x}{{dx}}=\int\limits_{0}^{t}{{g\left( t \right)dt}}

or               x = x_o+ \int\limits_{0}^{t}{{g\left( t \right)dt}}.            This gives x as a function of time.

Illustration

The acceleration of a particle is given by a = 4t – 30, where a is in m/s² and t is in s. Determine the velocity and displacement as functions of time. The initial displacement at  t = 0 is –5m, and the initial velocity is 3 m/s.

Solution

a = \displaystyle \frac{{dv}}{{dt}} = 4t – 30

dv= (4t – 30)dt

\displaystyle \int\limits_{3}^{v}{{dv}}=\int\limits_{0}^{t}{{\left( {4t-30} \right)dt}}

Þ     v – 3 = 2t² – 30 t

or      v = 3 – 30t + 2t²   m/s

Now v = \displaystyle \frac{{dx}}{{dt}} = 3 – 30 t + 2t²

dx= (3 – 30t + 2t²)dt

\displaystyle \int\limits_{{-5}}^{x}{{dx}}=\int\limits_{0}^{t}{{\left( {3-30t+2{{t}^{2}}} \right)dt}}

x + 5 = 3t – 15t² + \frac{2}{3}t ³

or      x = -5 + 3t – 15t² + \frac{2}{3}t ³  m

Acceleration  as a Function of position i.e. a = f(x)

Here         a = \displaystyle v\frac{{dv}}{{dx}}=f\left( x \right)

\displaystyle \int\limits_{u}^{v}{{vdv}}=\int\limits_{{{{x}_{o}}}}^{x}{{f\left( x \right)dx}}

or      v²   =  u² + 2 \displaystyle \int\limits_{{{{x}_{o}}}}^{x}{{f\left( x \right)dx}}

This gives v = g(x), a function of x. Now we can substitute  \displaystyle \frac{{dx}}{{dt}}=v, separate variables, and integrate

\displaystyle \int\limits_{{{{x}_{o}}}}^{x}{{\frac{{dx}}{{g\left( x \right)}}=\int\limits_{0}^{t}{{dt}}}}                or         t = \displaystyle \int\limits_{{{{x}_{o}}}}^{x}{{\frac{{dx}}{{g\left( x \right)}}=\int\limits_{0}^{t}{{dt}}}}

Which gives t as a function of x. Finally, we can rearrange to get x as a function of t.

Illustration

The relation between time t and distance x is t = ax² + bx where a and b are constant. The acceleration is

(a)  – 2abv²                  

(b)    2 bv³                                 

(c)  –2 av³     

(d)   2 av²

Solution

(c)     t = ax² + bx                         or        \displaystyle \frac{{dt}}{{dx}}=2ax+b

or      \displaystyle v=\frac{{dx}}{{dt}}=\frac{1}{{2ax+b}}

differentiating again w.r.t. time t, we get acceleration

            a=  \displaystyle \frac{{dv}}{{dt}}=\frac{{-2a}}{{{{{(2ax+b)}}^{2}}}}\frac{{dx}}{{dt}}=\frac{{-2a}}{{{{{(2ax+b)}}^{3}}}}=-2a\,{{v}^{3}}.

Acceleration  given as a Function of Velocity, a = f(v)

Here a = \frac{{dv}}{{dt}}=f\left( v \right)

\displaystyle \int\limits_{0}^{t}{{dt}}=\int\limits_{u}^{v}{{\frac{{dv}}{{f\left( v \right)}}}}

This gives t as a function of v. Then it would be necessary to solve for v as a function of t so that equation v = \frac{{dx}}{{dt}} can be integrated to obtain position coordinate x as function of time t.

Alternatively,

a = \displaystyle v\frac{{dv}}{{dx}}=f\left( v \right)

\displaystyle \int\limits_{u}^{v}{{\frac{{vdv}}{{f\left( v \right)}}=\int\limits_{{{{x}_{o}}}}^{x}{{dx}}}}

or      x = x_o+ \displaystyle \int\limits_{u}^{v}{{\frac{{vdv}}{{f\left( v \right)}}}}

This equation gives x in terms of v without explicit reference to t.

Illustration

A particle moves in a straight line with deceleration whose modulus depends on the velocity v of the particle as a = a \displaystyle \sqrt{v},  where a is a positive constant. The initial velocity of the particle is v_o. What distance will it traverse before it stops. What time will it take to cover that distance?

Solution

Here   a = \displaystyle \frac{{dv}}{{dt}} = –av^1/2

\displaystyle \int\limits_{{{{v}_{o}}}}^{v}{{{{v}^{{-1/2}}}}}dv=-\alpha \int\limits_{0}^{t}{{dt}}                                   

or         2(v^1/2 – v₀^1/2) = –at                                        (1)

Let the particle come to rest at time T. Then at t = T, v = 0.

Equation (1) gives 2(0 – v₀^1/2) = –aT

or         T = 2 \sqrt{{{{v}_{o}}}}/\alpha

Distance covered   s = \displaystyle \int\limits_{0}^{T}{{vdt}}=\int\limits_{0}^{T}{{\left( {v_{0}^{{1/2}}-\frac{1}{2}\alpha t} \right)dt=\frac{{2v_{{^{0}}}^{{3/2}}}}{{3\alpha }}}}

Illustration

The acceleration of a body is related to its velocity as a = kv². If at x=0 the velocity of the body is 5m/sec. Find the relation of velocity with time t.

Solution

given \displaystyle a=k{{v}^{2}}

\displaystyle \Rightarrow \text{  }\frac{{dv}}{{dt}}=k{{v}^{2}}

\displaystyle \Rightarrow \text{  }\frac{{dv}}{{{{v}^{2}}}}=kdt

\displaystyle \Rightarrow \text{  }\int\limits_{5}^{v}{{\frac{{dv}}{{{{v}^{2}}}}=k\int\limits_{0}^{t}{{dt}}}}

\displaystyle \Rightarrow \text{  }\left[ {\frac{{{{v}^{{-2+1}}}}}{{-2+1}}} \right]_{5}^{v}=k\left[ t \right]_{0}^{t}

\displaystyle \Rightarrow \text{  -}\left[ {\frac{1}{v}} \right]_{5}^{v}=kt

\displaystyle \Rightarrow \text{  }\left[ {\frac{1}{v}-\frac{1}{5}} \right]=-kt

\displaystyle \Rightarrow \text{  }\frac{1}{v}=-kt+\frac{1}{5}

\displaystyle \Rightarrow \text{  }v=\frac{5}{{1-5kt}}

Remember that in the motion with variable acceleration, the graphs between different quantities depends on their mutual relation based on the situation given in the questions