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.

#### 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, $$a=\dfrac{dv}{dt}$$

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

$$=\dfrac{d^2x}{dt^2}$$

#### 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 $$t$$

#### 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

#### 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

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

#### 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.

#### 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

#### 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.

#### 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 $$x$$.

#### 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)

#### 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

$$a=v\,\dfrac{dv}{dx}$$

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

#### 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

$$a=v\,\dfrac{dv}{dx}$$

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

#### 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