Non-Uniform Motion With Non-Uniform Acceleration (Board 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  changesat 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 .

Hence there are three types of question based on relations of acceleration

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

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

(iii) Acceleration  given as a Function of Velocity i.e.  a = f(v)

Illustration based on (i)

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

Illustration based on (ii)

The acceleration of a body is related to its position as a = kx². If at x=0 the velocity of the body is 4m/sec. Find the relation of velocity with position x.

Solution

Given   \displaystyle a=k{{x}^{2}}

hence  \displaystyle \frac{{dv}}{{dt}}=k{{x}^{2}}

or  \displaystyle \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}=k{{x}^{2}}

or  \displaystyle \frac{{dv}}{{dx}}.v=k{{x}^{2}}

or  \displaystyle vdv=k{{x}^{2}}dx

or \displaystyle \int\limits_{4}^{v}{{vdv}}=k\int\limits_{0}^{x}{{{{x}^{2}}dx}}

or  \displaystyle \left[ {\frac{{{{v}^{2}}}}{2}} \right]_{4}^{v}=k\left[ {\frac{{{{x}^{3}}}}{3}} \right]_{0}^{x}

or \displaystyle \frac{{{{v}^{2}}}}{2}-\frac{{{{4}^{2}}}}{2}=k\left[ {\frac{{{{x}^{3}}}}{3}-0} \right]

or  \displaystyle {{v}^{2}}=2k\left[ {\frac{{{{x}^{3}}}}{3}} \right]+16

or \displaystyle v=\sqrt{{2k\left[ {\frac{{{{x}^{3}}}}{3}} \right]+16}}

Illustration based on (iii)

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