Informative line

Non Uniform Acceleration

Learn interpretation of variable acceleration, practice to calculation of velocity and acceleration, position of particle as a function of time & velocity when acceleration is given in terms of displacement.

Interpretation of Variable Acceleration

  • When rate of change of velocity of a moving particle varies with time, then the acceleration is called variable acceleration.
  • A body is said to be moving with variable acceleration, if its velocity changes by unequal amount in equal intervals of time.
  • Mathematically, if acceleration \('a'\) is a function of time or changes with time, then it is called variable acceleration.

Illustration Questions

Which one of the following is constant acceleration?

A \(a=3\,t^2\)

B \(v=6\,t\)

C \(a=6\,t\)

D \(v=6\,t^2\)

×

Option (A) is incorrect because 'a' is a function of time. Thus, it is variable acceleration.

\(v=6\,t\)

\(\Rightarrow\;\dfrac{dv}{dt}=a=6\)

Option (B) is correct because 'a' is not a function of time. Thus, it is constant acceleration.

Option (C) is incorrect because 'a' is a function of time. Thus, it is variable acceleration.

\(v=6\,t^2\)

\(\dfrac{dv}{dt}=a=12\,t\)

Option (D) is incorrect because 'a' is a function of time. Thus, it is variable acceleration.

Which one of the following is constant acceleration?

A

\(a=3\,t^2\)

.

B

\(v=6\,t\)

C

\(a=6\,t\)

D

\(v=6\,t^2\)

Option B is Correct

Calculation of Velocity and Acceleration as a Function of Time

  • The velocity of a particle moving along a straight line can be obtained when the position of the particle is given as a function of time.
  • The velocity of the particle is equal to the derivative of the position of the particle with respect to time.
  • Mathematically, if \(x=f(t)\)

                  then velocity, \(v=\dfrac{dx}{dt}\)

  • Similarly, acceleration of the particle is equal to the derivative of the velocity with respect to time.
  • If, \(v=\dfrac{dx}{dt}\)  then acceleration, \(a=\dfrac{dv}{dt}\)

\(=\dfrac{d}{dt}\,\left[\dfrac{dx}{dt}\right]\)

\(=\dfrac{d^2x}{dt^2}\)

Illustration Questions

The position of a particle, moving along a straight line is given by the equation \(x=5\,t^3+3\,t-2\), with \(x\) in meters and \(t\) in seconds as a function of time. Determine acceleration and velocity of the particle.

A \(15t^2+3,\;30t\)

B \(16\,t^3+5,\;25t\)

C \(15t^3+3t,\;30t+3\)

D \(5t^2+3,\;15t\)

×

The position of particle

\(x=5t^3+3t-2\)

Velocity of the particle, \(v=\dfrac{dx}{dt}\)

\(v=\dfrac{d}{dt}(5t^3+3t-2)\)

\(v=15t^2+3\)

Acceleration of particle,

\(a=\dfrac{dv}{dt}\)

\(=\dfrac{d}{dt}(15t^2+3)\)

\(a=30\,t\)

The position of a particle, moving along a straight line is given by the equation \(x=5\,t^3+3\,t-2\), with \(x\) in meters and \(t\) in seconds as a function of time. Determine acceleration and velocity of the particle.

A

\(15t^2+3,\;30t\)

.

B

\(16\,t^3+5,\;25t\)

C

\(15t^3+3t,\;30t+3\)

D

\(5t^2+3,\;15t\)

Option A is Correct

Calculation of Time and Distance at the Instant when Particle Changes its Direction

  • The velocity of a particle moving along a straight line becomes zero at the instant at which it changes its direction.
  • To calculate the time and the position of the particle at the instant when it changes its direction, following steps are to be taken 
  1. As,  \(v=f(t)\)
  2. Put  \(f(t)=v=0\) and calculate time  \(t\)
  3. \(x=x_0+\int\limits^t_0\,v\,dt\)
  4. Find \(x\) by putting value of \(t\)

Illustration Questions

The velocity of a particle moving along a straight line, is given by the equation \(v=3t-2\) and the initial position of the particle is \(4\,m\) . Determine the time and the position of the particle when it changes its direction.

A \(\dfrac{2}{3}\,sec\,,\;\dfrac{10}{3}\,m\)

B \(\dfrac{2}{5}\,sec\;,\,\dfrac{5}{3}\,m\)

C \(\dfrac{5}{6}\,sec\;,\,\dfrac{12}{5}\,m\)

D \(\dfrac{4}{3}\,sec\,,7\,m\)

×

Given :

velocity, \(v=3t-2\)

Initial position, \(x_0=4\,m\)

The instant at which particle changes its direction, the velocity becomes zero.

\(v=0\)

\(\Rightarrow\;3t-2=0\)

\(t=\dfrac{2}{3}sec\)

Since, \(x=x_0+ \int \limits ^t_0\,v\,dt\)

\(x=4+ \int \limits ^{2/3}_0\,(3t-2)\,dt\)

\(x=4+\left[\dfrac{3t^2}{2}-2t\right]^{2/3}_0\)

At  \(t=\dfrac23 sec\)

\(x=4+\dfrac{3}{2}×\left(\dfrac{2}{3}\right)^2-2×\dfrac{2}{3}-0\)

\(=4+\dfrac{2}{3}-\dfrac{4}{3}\)

\(=\dfrac{10}{3}\;m\)

The velocity of a particle moving along a straight line, is given by the equation \(v=3t-2\) and the initial position of the particle is \(4\,m\) . Determine the time and the position of the particle when it changes its direction.

A

\(\dfrac{2}{3}\,sec\,,\;\dfrac{10}{3}\,m\)

.

B

\(\dfrac{2}{5}\,sec\;,\,\dfrac{5}{3}\,m\)

C

\(\dfrac{5}{6}\,sec\;,\,\dfrac{12}{5}\,m\)

D

\(\dfrac{4}{3}\,sec\,,7\,m\)

Option A is Correct

Illustration Questions

A particle is moving along a straight line with acceleration \(a=3t-3\). Determine the position of the particle and the time when acceleration is zero. Consider the initial velocity and the initial position of the particle to be zero.

A \(-1\,m\,,\,1\,sec\)

B \(-2\,m,\;1\,sec\)

C \(-3\,m,\;1\,sec\)

D \(-4\,m,\;1\,sec\)

×

Acceleration \(a=3t-3\)

Initial velocity \(=0\)

Initial position of particle \(=0\)

Calculation of time when acceleration is zero,

Given, \(a=3t-3\)

Putting \(a=0\)

\(0=3t-3\)

\(t=1\,sec\)

Calculation of velocity

Since,   \(\int\limits ^v_0dv=\int\limits ^t_0a\,dt\)

\(v=\int\limits ^t_0(3t-3)\,dt\)

\(v=\dfrac{3t^2}{2}-3t\)

Calculation of position

Since,   \(\int\limits ^x_0dx=\int\limits ^t_0vdt\)

\(x=\displaystyle\int\limits ^t_0\left(\dfrac{3t^2}{2}-3t\right)dt\)

\(=\dfrac{t^3}{2}-\dfrac{3}{2}t^2\)

As \(t=1\,sec\), is the time when acceleration of the particle becomes zero.

So, to calculate initial position, put  \(t=1\,sec\)  in equation of \(x\).

\(\therefore\;x=\dfrac{t^3}{2}-\dfrac{3}{2}t^2\)

\(x=\dfrac{1}{2}-\dfrac{3}{2}×1\)

\(x=-1\,m\)

A particle is moving along a straight line with acceleration \(a=3t-3\). Determine the position of the particle and the time when acceleration is zero. Consider the initial velocity and the initial position of the particle to be zero.

A

\(-1\,m\,,\,1\,sec\)

.

B

\(-2\,m,\;1\,sec\)

C

\(-3\,m,\;1\,sec\)

D

\(-4\,m,\;1\,sec\)

Option A is Correct

Calculation of Position of Particle as a Function of Time

  • Consider a particle is moving along a straight line and its position is given as a function of time.
  • Then, the velocity of the particle is equal to the derivative of the position of the particle with respect to time.

Let the position of particle \(=x(t)\)

then velocity, \(v=\dfrac{d[x(t)]}{dt}\) ...(1)

  • Similarly, acceleration of the particle is equal to the derivative of the velocity with respect to time.

If \(v=\dfrac{d[x(t)]}{dt}\)

\(a=\dfrac{dv}{dt}=\dfrac{d^2}{dt}[x(t)]\)  ...(2)

  • On the other hand, if position is to be calculated and velocity is given as a function of time then following steps are to be taken 

As, \(v=\dfrac{dx}{dt}\)

\(dx=v\,dt\)

Now, integrating both the sides from initial to final conditions, we get 

\(\int\limits^x_{x_0}dx=\int\limits ^t_{t_0}v\,dt\)

Illustration Questions

A particle is moving along a straight line with velocity \(v=6t-2\). Calculate the position of the particle as a function of time.

A \(3t^2-2t+C\)

B \(5t^2-2t+C\)

C \(5t^2+2t+C\)

D \(5t^2+6t+C\)

×

The velocity of the particle moving along a straight line is given as 

\(v=6t-2\)

As, \(v=\dfrac{dx}{dt}\)

\(dx=v\,dt\)

Integrating both sides, 

\(\int\limits dx=\int v\,dt\)

\(x=\int (6t-2)\,dt\)

\(x=3t^2-2t+C\)

where \(C\) is an arbitrary constant which can be calculated if initial conditions of the particle are given.

A particle is moving along a straight line with velocity \(v=6t-2\). Calculate the position of the particle as a function of time.

A

\(3t^2-2t+C\)

.

B

\(5t^2-2t+C\)

C

\(5t^2+2t+C\)

D

\(5t^2+6t+C\)

Option A is Correct

Calculation of Position of Particle from Velocity at a given Instant

  • Velocity of a particle moving along a straight line is given as

\(v=\dfrac{dx}{dt}\)

where, \(x\) is the position of the particle.

  • When the velocity of the particle is given as the function of time, then the position of particle at a given instant can be calculated by :

As,  \(v=\dfrac{dx}{dt}\)

Integrating both sides

\(\int\limits ^t_0v\,dt=\int\limits^x_0dx\)

where, \(t\) is the time at which position of the particle is to be determined.

Illustration Questions

The velocity of a particle moving along a straight line is given as \(v=6\,t-2\). Calculate its position at time \(t=4\,sec\), if initial position of particle is zero. 

A \(20\,m\)

B \(30\,m\)

C \(40\,m\)

D \(50\,m\)

×

Given :

Velocity, \(v=6\,t-2\)

Initial position \(=0\,m\)

As, \(v=\dfrac{dx}{dt}=6\,t-2\)

\(dx=(6\,t-2)\,dt\)

Integrating both sides

\(\int\limits^x_0dx=\int\limits^t_0\,(6\,t-2)\,dt\)

\(x=3\,t^2-2\,t\)

To calculate the position of the particle, put  \(t=4\,sec\)  in equation of \(x\)

As, \(x=3\,t^2-2\,t\)

\(x=3(4)^2-2×4\)

\(=48-8\)

\(=40\,m\)

The velocity of a particle moving along a straight line is given as \(v=6\,t-2\). Calculate its position at time \(t=4\,sec\), if initial position of particle is zero. 

A

\(20\,m\)

.

B

\(30\,m\)

C

\(40\,m\)

D

\(50\,m\)

Option C is Correct

Illustration Questions

A particle is moving along a straight line with velocity \(v=8\,t+3\). Calculate the position of the particle as a function of time, if its initial position is \(3\,m\).

A \(4t^2+3t+3\)

B \(5t+9\)

C \(4t^2+9t+16\)

D \(8t+3\)

×

The velocity of particle moving along a straight line \(=8\,t+3\)

Initial position \(=3\,m\)

Using   \(\int\limits^x_{x_0}dx=\int\limits^t_0v\,dt\)

\(\int\limits^x_3dx=\int\limits^t_0\,(8\,t+3)\,dt\)

\([x]^x_3=\left[4\,t^2+3\,t\right]^t_0\)

\(x-3=4\,t^2+3\,t\)

\(x=4\,t^2+3\,t+3\)

A particle is moving along a straight line with velocity \(v=8\,t+3\). Calculate the position of the particle as a function of time, if its initial position is \(3\,m\).

A

\(4t^2+3t+3\)

.

B

\(5t+9\)

C

\(4t^2+9t+16\)

D

\(8t+3\)

Option A is Correct

Calculation of Acceleration from Velocity

  • Consider a particle, moving along a straight line with velocity \(v\) such that \(v\) is a function of \(t\).

\(\therefore v=f(t)\)

  • Then, the acceleration of the particle is given as

\(a=\dfrac{dv}{dt}\)

where, \(v=\) velocity of the particle

  • This is the instantaneous acceleration.

Illustration Questions

Calculate the acceleration of a particle which is moving along a straight line with velocity \(v=3\,t^2+2\)  at  \(t=1\,sec\).

A \(4\,m/sec^2\)

B \(3\,m/sec^2\)

C \(6\,m/sec^2\)

D \(2\,m/sec^2\)

×

The acceleration of the particle is given as

\(a=\dfrac{dv}{dt}\)

where, \(v=\) velocity of the particle

Given : velocity, \(v=3\,t^2+2,\;t=1\,sec\)

acceleration, \(a=\dfrac{d}{dt}(3\,t^2+2)\)

\(a=6\,t\)

At  \(t=1\,sec\)

\(a=6×1\)

\(a=6\,m/sec^2\)

Calculate the acceleration of a particle which is moving along a straight line with velocity \(v=3\,t^2+2\)  at  \(t=1\,sec\).

A

\(4\,m/sec^2\)

.

B

\(3\,m/sec^2\)

C

\(6\,m/sec^2\)

D

\(2\,m/sec^2\)

Option C is Correct

Acceleration as a Function of Distance \((x)\)

  • A particle is moving with velocity \(v\), along a straight line given by

\(v=\dfrac{dx}{dt}\)

  • As acceleration of a particle is given by 

\(a=\dfrac{dv}{dt}\)

multiplying by \(dx\) at both the sides 

\(a.dx=\dfrac{dx}{dt}.dv\)

\(a\,dx=v\,dv\)

\(a=v\,\dfrac{dv}{dx}\)  ...(1)

Equation (1) can be described as :

Acceleration as a function of position \(x\).

Illustration Questions

The velocity of a particle moving along a straight line is given as \(v=3x^2+5x\). Determine the acceleration of the particle as a function of \(x\).

A \(18x^3+45x^2+25x\)

B \(16x^3+17x^2+20x\)

C \(16x^2+5x+7\)

D \(11x^3+5x\)

×

The velocity of the particle is given by

\(v=3x^2+5x\)

As acceleration

\(a=v\,\dfrac{dv}{dx}\)

\(\dfrac{dv}{dx}=\dfrac{d}{dx}\left[3x^2+5x\right]\)

\(\dfrac{dv}{dx}=6x+5\)

and \(v=3x^2+5x\)

So,  \(v\,\dfrac{dv}{dx}=(3x^2+5x)\,(6x+5)\)

\(a=18x^3+15x^2+30x^2+25x\)

\(a=18x^3+45x^2+25x\)

The velocity of a particle moving along a straight line is given as \(v=3x^2+5x\). Determine the acceleration of the particle as a function of \(x\).

A

\(18x^3+45x^2+25x\)

.

B

\(16x^3+17x^2+20x\)

C

\(16x^2+5x+7\)

D

\(11x^3+5x\)

Option A is Correct

Acceleration of a Particle at a Given Distance 

  • Let velocity of a particle be given as a function of \(x\).
  • As acceleration of particle given by

\(a=\dfrac{dv}{dt}\)

multiplying both the sides by \(dx\)

\(a\,dx=\dfrac{dx}{dt}.\,dv=v.dv\)

\(a=v\,\dfrac{dv}{dx}\) ...(1)

Illustration Questions

The velocity of a particle moving along a straight line is given as \(v=3x^2+5x\). When particle is at \(1\,m\), determine the acceleration of the particle.

A \(50\,m/s^2\)

B \(40\,m/s^2\)

C \(88\,m/s^2\)

D \(60\,m/s^2\)

×

The velocity of the particle, 

\(v=3x^2+5x\)

As  \(v=3x^2+5x\)

\(\dfrac{dv}{dx}=6x+5\)

\(a=v\,\dfrac{dv}{dx}\)

\(a=(3x^2+5x)\,(6x+5)\)

Put,  \(x=1\,m\)  to find acceleration at \(1\,m\)

\(a=(3×1^2+5×1)\,(6×1+5)\)

\(a=88\,m/s^2\)

The velocity of a particle moving along a straight line is given as \(v=3x^2+5x\). When particle is at \(1\,m\), determine the acceleration of the particle.

A

\(50\,m/s^2\)

.

B

\(40\,m/s^2\)

C

\(88\,m/s^2\)

D

\(60\,m/s^2\)

Option C is Correct

Calculation of Velocity when Acceleration is given in terms of Displacement

  • The acceleration of particle is given as

\(a=v\,\dfrac{dv}{dx}\)

\(\int \limits^x_{x_0}a\,dx=\int \limits^v_{v_0}v\,dv\)

Illustration Questions

Acceleration of a particle moving along a straight line is given as \(a=3x+2\). If velocity of the particle is \(0\,m/s\) at the origin, then calculate the velocity as a function of  \(x\).

A \(3x^2+4x\)

B \(\sqrt{3x^2+4x}\)

C \(x+5\)

D \(2x+5\)

×

The acceleration of the particle, \(a=3x+2\)

The velocity of the particle at the origin \(=0\,m/s\)

\(\therefore\;\int \limits^x_0a\,dx=\int \limits^v_0\,v\,dv\)

\(\int \limits^x_0\,(3x+2)\,dx=\dfrac{v^2}{2}\)

\(\dfrac{3x^2}{2}+2x=\dfrac{v^2}{2}\)

\(\sqrt{3x^2+4x}=v\)

Acceleration of a particle moving along a straight line is given as \(a=3x+2\). If velocity of the particle is \(0\,m/s\) at the origin, then calculate the velocity as a function of  \(x\).

A

\(3x^2+4x\)

.

B

\(\sqrt{3x^2+4x}\)

C

\(x+5\)

D

\(2x+5\)

Option B is Correct

Calculation of Velocity at a Given Distance

  • Acceleration of particle is given as

\(a=v\,\dfrac{dv}{dx}\)

\(\int\limits^x_{x_0}a\,dx=\int\limits^v_{v_0}v\,dv\)

Illustration Questions

The acceleration of a particle moving along a straight line is given as \(a=4x+3\). If the velocity of the particle at origin is \(0\,m/sec\), then calculate its velocity at \(x=3m.\)

A \(54\,m/s\)

B \(\sqrt{27}\,m/s\)

C \(\sqrt{54}\,m/s\)

D \(\sqrt{91}\,m/s\)

×

The acceleration of particle  \(a=4x+3\)

Velocity at origin \(=0\,m/sec\)

Position of particle, \((x)=3\,m\)

 

As   \(\int\limits^v_0v\,dv=\int\limits^x_0a\,dx\)

\(\left[\dfrac{v^2}{2}\right]^v_0=\int\limits^3_0\,(4x+3)\,dx\)

\(\dfrac{v^2}{2}=\left[2x^2+3x\right]^3_0\)

\(\dfrac{v^2}{2}=27\)

\(v^2=54\)

\(v=\sqrt{54}\,m/s\)

The acceleration of a particle moving along a straight line is given as \(a=4x+3\). If the velocity of the particle at origin is \(0\,m/sec\), then calculate its velocity at \(x=3m.\)

A

\(54\,m/s\)

.

B

\(\sqrt{27}\,m/s\)

C

\(\sqrt{54}\,m/s\)

D

\(\sqrt{91}\,m/s\)

Option C is Correct

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