Informative line

### Uniform Acceleration

Practice equation to calculate of maximum height attained by a ball and velocity and displacement of a particle moving along a straight line with constant acceleration. Learn uniform and constant acceleration formula.

# Acceleration

• Acceleration is defined as the rate of change of velocity with respect to time.

$$\vec a_{avg}=\dfrac{\Delta \vec v}{\Delta \,t}$$

• From above relation, it is clear that acceleration is also a vector quantity.

Note :

1. Direction of acceleration vector is same as that of change in velocity vector.
2. There are two ways in which velocity can change.

(a) Change in magnitude

(b) Change in direction

• So, two types of accelerations can be produced.

## Acceleration in one-dimensional motion

• In one-dimensional motion, acceleration can have only two directions, either positive or negative.
• To understand one-dimensional motion with constant acceleration, consider a particle is moving along a straight line with uniform (constant) acceleration $$'a'$$. If the velocity of the particle at $$t=t_1$$ is  $$v_0$$  and at  $$t=t_2$$  is $$v$$, then

$$v=v_0+a(t_2-t_1)$$

$$v=v_0+a(\Delta\,t)$$

• It is clear from the above equation that, if

(a) Acceleration and velocity are in same direction, magnitude of velocity ,i.e., speed of the particle will increase.

(b) Acceleration and velocity are in opposite direction, magnitude of velocity ,i.e., speed of the particle will decrease.

#### A particle is moving along a straight line. Choose the correct option regarding the speed of particle.

A velocity is  $$3\,m/sec$$  and acceleration is $$-5\,m/sec^2$$, speed is increasing

B velocity is  $$3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$, speed is increasing

C velocity is  $$-3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$, speed is increasing

D velocity is  $$3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$ , speed is decreasing

×

Since, the direction of velocity and acceleration is different so, speed is decreasing.

Hence, option (A) is incorrect.

Since, the direction of velocity and acceleration is same so, speed is increasing.

Hence, option (B) is correct.

Since, the direction of velocity and acceleration is different so, speed is decreasing.

Hence, option (C) is incorrect.

Since, the direction of velocity and acceleration is same so, speed is increasing.

Hence, option (D) is incorrect.

### A particle is moving along a straight line. Choose the correct option regarding the speed of particle.

A

velocity is  $$3\,m/sec$$  and acceleration is $$-5\,m/sec^2$$, speed is increasing

.

B

velocity is  $$3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$, speed is increasing

C

velocity is  $$-3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$, speed is increasing

D

velocity is  $$3\,m/sec$$  and acceleration is  $$5\,m/sec^2$$ , speed is decreasing

Option B is Correct

# Calculation of Velocity and Displacement of a Particle Moving along a Straight Line with Constant Acceleration

• Let us consider a particle, moving along a straight line with constant acceleration $$'a'$$.
• If the initial velocity of the particle at  $$t=t_0$$  is  $$v_0$$ , then the velocity at time $$t$$ will be

$$a=\dfrac{v-v_0}{t-t_0}$$

$$\Rightarrow\;v-v_0=a(t-t_0)$$

$$\Rightarrow\;v=v_0+a(t-t_0)$$

$$\Rightarrow\;v=v_0+a(\Delta t)$$

• Since, velocity of the particle is varying uniformly, we can say

$$v_{avg}=\dfrac{v+v_0}{2}$$

• Average velocity $$(v_{avg})$$ is given as

$$v_{avg}=\dfrac{\text{Displacement}}{\text{Time interval}}=\dfrac{\Delta x}{\Delta t}$$

$$\Rightarrow\;\Delta x=v_{avg}(\Delta t)$$

$$\Rightarrow\;\Delta x=\left(\dfrac{v+v_0}{2}\right)\,\Delta t$$

Using  $$v=v_0+a(\Delta t)$$

$$\Rightarrow\;\Delta x=v_0(\Delta t)+\dfrac{1}{2}a(\Delta t)^2$$

#### A particle is moving along a straight line with initial velocity $$-3\,m/sec$$  and acceleration $$8\,m/sec^2$$. Calculate velocity and displacement of the particle after $$8\,seconds.$$

A $$61\,m/sec,\,232\,m$$

B $$6\,m/sec,\,13\,m$$

C $$68\,m/sec,\,120\,m$$

D $$50\,m/sec,\,60\,m$$

×

Initial velocity of the particle, $$u=-3\,m/s$$

Acceleration, $$a=8\,m/s^2$$

Time duration, $$\Delta t=8\,sec$$

Velocity after $$8\,sec$$

$$v=u+at$$

$$v=-3+8\,(8)$$

$$v=61\,m/s$$

Displacement after $$8\,sec$$

$$\Delta x=ut+\dfrac{1}{2}at^2$$

$$\Delta x=-3\,(8)+\dfrac{1}{2}8\,(8)^2$$

$$\Delta x=232\ m$$

### A particle is moving along a straight line with initial velocity $$-3\,m/sec$$  and acceleration $$8\,m/sec^2$$. Calculate velocity and displacement of the particle after $$8\,seconds.$$

A

$$61\,m/sec,\,232\,m$$

.

B

$$6\,m/sec,\,13\,m$$

C

$$68\,m/sec,\,120\,m$$

D

$$50\,m/sec,\,60\,m$$

Option A is Correct

#### Consider a particle is moving along a straight line. If its displacement is $$14\ m$$ in first $$2\ sec$$ and $$22\ m$$ in the next $$2\ sec$$ then calculate the initial velocity and acceleration of the particle.

A $$5\,m/s\;,\,2\,m/s^2$$

B $$3\,m/s\;,\,8\,m/s^2$$

C $$2\,m/s\;,\,6\,m/s^2$$

D $$6\,m/s\;,\,2\,m/s^2$$

×

Let initial velocity be $$v_0$$  and acceleration be $$a$$.

Displacement in first $$2\ sec=14\ m$$

Using $$ut+\dfrac{1}{2}at^2=s$$

$$\Rightarrow\;v_0(2)+\dfrac{1}{2}a(2)^2=14$$  ...(1)

Displacement in next  $$2\,sec=22\,m$$

Total displacement in  $$4\,sec=22+14$$

$$=36\,m$$

Displacement in $$4\,sec=36\,m$$

$$\Rightarrow\;v_0(4)+\dfrac{1}{2}a(4)^2=36$$ ...(2)

Solving equation (1) and (2) we get,

$$v_0=5\,m/s$$

$$a=2\,m/s^2$$

### Consider a particle is moving along a straight line. If its displacement is $$14\ m$$ in first $$2\ sec$$ and $$22\ m$$ in the next $$2\ sec$$ then calculate the initial velocity and acceleration of the particle.

A

$$5\,m/s\;,\,2\,m/s^2$$

.

B

$$3\,m/s\;,\,8\,m/s^2$$

C

$$2\,m/s\;,\,6\,m/s^2$$

D

$$6\,m/s\;,\,2\,m/s^2$$

Option A is Correct

# Displacement in the Given Time Interval

• Let us consider a particle, moving along a straight line with constant acceleration $$'a'$$. If the initial velocity of the particle at $$t=t_0$$ is $$v_0$$, then the velocity at time $$t$$ will be

$$a=\dfrac{v-v_0}{t-t_0}$$

$$\Rightarrow\;v-v_0=a(t-t_0)$$

$$\Rightarrow\;v=v_0+a(t-t_0)$$

$$\Rightarrow\;v=v_0+a(\Delta t)$$

• Since, velocity of the particle is varying uniformly, we can say

$$v_{avg}=\dfrac{v+v_0}{2}$$

• Average velocity $$(v_{avg})$$ is given as

$$v_{avg}=\dfrac{\text{Displacement}}{\text{Time interval}}=\dfrac{\Delta x}{\Delta t}$$

$$\Rightarrow\;\Delta x=v_{avg}(\Delta t)$$

$$\Rightarrow\;\Delta x=\left(\dfrac{v+v_0}{2}\right)\,\Delta t$$

Using  $$v=v_0+a(\Delta t)$$

$$\Rightarrow\;\Delta x=v_0(\Delta t)+\dfrac{1}{2}a(\Delta t)^2$$

where, $$\Delta x=x_2-x_1$$

#### A particle having a constant acceleration of $$2\ m/s^2$$ and initial velocity $$5\,m/s$$ moves in a straight line for $$8\,sec$$. Find its displacement in the last two seconds.

A $$16\ m$$

B $$25\ m$$

C $$20\ m$$

D $$38\ m$$

×

Initial velocity, $$u=5\ m/s$$

Acceleration, $$a=2\ m/s^2$$

Time, $$t=8\ sec$$

Displacement in first $$6\,seconds$$

$$x_1=ut+\dfrac{1}{2}\,at^2$$

$$x_1=5\,(6)+\dfrac{1}{2}(2)(6)^2$$

$$x_1=66\,m$$

Displacement in $$8\,seconds$$

$$x_2=u\,t+\dfrac{1}{2}\,a\,t^2$$

$$x_2=5\,(8)+\dfrac{1}{2}\,(2)\,(8)^2$$

$$x_2=104\,m$$

Displacement in last two seconds

$$=x_2-x_1$$

$$=104-66$$

$$=38\,m$$

### A particle having a constant acceleration of $$2\ m/s^2$$ and initial velocity $$5\,m/s$$ moves in a straight line for $$8\,sec$$. Find its displacement in the last two seconds.

A

$$16\ m$$

.

B

$$25\ m$$

C

$$20\ m$$

D

$$38\ m$$

Option D is Correct

# Displacement of the Particle

• Consider a particle with uniform acceleration $$'a\,'$$ is moving along a straight line.
• If the velocity of the particle changes from $$v_0$$ to $$v$$ in a time interval of $$\Delta t$$, the displacement in this time interval can be obtained by,

$$\Delta x=v_o(\Delta t)+\dfrac{1}{2}a(\Delta t)^2$$  .....(1)

where $$\Delta t=\dfrac{v-v_0}{a}$$ .....(2)

• After eliminating $$\Delta t$$ from equation (1), we get

$$v^2=v_0^2+2a(\Delta x)$$

$$\Delta x=\dfrac{v^2-v_0^2}{2\,a}$$

where ,  $$v_0=$$ Initial velocity

$$v=$$ Final velocity

$$a=$$ Acceleration

$$\Delta x=$$ Displacement

#### A particle moving with a constant acceleration of $$4\,m/s^2$$, changes its velocity from $$5\,m/s$$  to  $$13\,m/s$$ in a certain time interval. Calculate the displacement of the particle in this time interval.

A $$3\,m$$

B $$18\,m$$

C $$2\,m$$

D $$4\,m$$

×

Initial velocity, $$v_0=5\ m/s$$

Final velocity, $$v=13\ m/s$$

Acceleration, $$a=4\ m/s^2$$

Displacement $$(\Delta x)=\dfrac{v^2-v_0^2}{2\,a}$$

$$\Delta x=\dfrac{(13)^2-(5)^2}{2×4}$$

$$\Delta x=\dfrac{144}{2\times4}$$

$$=\dfrac{72}{4}=18\,m$$

### A particle moving with a constant acceleration of $$4\,m/s^2$$, changes its velocity from $$5\,m/s$$  to  $$13\,m/s$$ in a certain time interval. Calculate the displacement of the particle in this time interval.

A

$$3\,m$$

.

B

$$18\,m$$

C

$$2\,m$$

D

$$4\,m$$

Option B is Correct

#### A particle is moving along a straight line with velocity $$18\,m/s$$. Acceleration of particle is $$-3\,m/s^2$$. Calculate the distance covered by the particle before it stops for the first time.

A $$20\ m$$

B $$54\ m$$

C $$37\ m$$

D $$23\ m$$

×

Initial velocity $$(u)=18\,m/s$$

Final velocity $$(v)=0$$

Acceleration $$(a)=-3\,m/s^2$$

Let displacement $$=x$$

$$v^2=u^2+2\,a\,x$$

$$0=(18)^2-2×3 ×x$$

$$x=54\,m$$

Distance $$=|x|$$

$$=54\ m$$

### A particle is moving along a straight line with velocity $$18\,m/s$$. Acceleration of particle is $$-3\,m/s^2$$. Calculate the distance covered by the particle before it stops for the first time.

A

$$20\ m$$

.

B

$$54\ m$$

C

$$37\ m$$

D

$$23\ m$$

Option B is Correct

# Maximum Height Attained by a Ball

• Various parameters of a particle projected vertically, comes under the domain of one dimensional motion as velocity of the particle is always along a straight line and its acceleration is along the line of motion.
• So, all the formulae of one-dimensional motion, are applicable for calculating the parameters.

## Calculation of maximum height

• Assuming, the motion in upward direction is positive and the motion in downward direction is negative.
• Let, maximum height attained by the particle when projected vertically upward be $$H.$$
• At height $$H$$, the velocity will be zero, i.e. $$v=0$$.
• Let the initial velocity of the particle is $$v_0$$ and the acceleration will be $$-g$$.
• So using

$$v^2=v_o^2+2a\Delta x$$      { here, v0 = initial velocity , v = final velocity }

$$0=v_o^2-2gH$$

$$H_{max}=\dfrac{v_o^2}{2g}$$

#### A ball is projected vertically upward with a speed of $$10\,m/s$$. What will be the maximum height attained by the ball?

A $$6\,m$$

B $$8\,m$$

C $$5\,m$$

D $$9\,m$$

×

Assuming upward direction as positive.

Initial velocity of particle,  $$v_o=10\,m/s$$

Using $$H_{max}=\dfrac{v_o^2}{2g}$$

$$H_{max}=\dfrac{10^2}{2×10}$$

$$=5\,m$$

### A ball is projected vertically upward with a speed of $$10\,m/s$$. What will be the maximum height attained by the ball?

A

$$6\,m$$

.

B

$$8\,m$$

C

$$5\,m$$

D

$$9\,m$$

Option C is Correct

#### A particle starting from rest, moves with an acceleration of $$8\,m/s^2$$ for first $$3\,sec$$ and then with $$2\,m/s^2$$ for next $$3\,sec$$. What will be the total displacement of the particle?

A $$110\,m$$

B $$117\,m$$

C $$120\,m$$

D $$130\,m$$

×

Event 1- For first  $$3\,sec$$

Let  $$\Delta t_1=3\,sec$$

Initial velocity, $$v_0=0$$

Acceleration, $$a_1=8\,m/s^2$$

Final velocity $$=v$$

using $$v=v_0+a\,t$$

$$=0+8×3$$

$$=24\,m/s$$

Displacement,

using $$\Delta x=v_0\,t+\dfrac{1}{2}\,a\,t^2$$

$$\Delta x_1=v_0\, \Delta t_1+\dfrac{1}{2}\,a_1×(\Delta t_1)^2$$

$$=0+\dfrac{1}{2}×8×3^2$$

$$=36\,m$$

Event 2- For the next  $$3\,sec$$

Initial velocity $$v=v_0'=24\,m/s$$

$$\Delta t_2=3\,sec,\;a_2=2\,m/s^2$$

Displacement,

Using $$\Delta x=v_0\,t+\dfrac{1}{2}\,a\,t^2$$

$$\Delta x_2=v_0\,'\, \Delta t_2+\dfrac{1}{2}\,a_2×\Delta t_2^2$$

$$=24×3+\dfrac{1}{2}×2×3^2$$

$$=81\,m$$

Total displacement

$$\Delta x=\Delta x_1+\Delta x_2$$

$$=36+81$$

$$=117\,m$$

### A particle starting from rest, moves with an acceleration of $$8\,m/s^2$$ for first $$3\,sec$$ and then with $$2\,m/s^2$$ for next $$3\,sec$$. What will be the total displacement of the particle?

A

$$110\,m$$

.

B

$$117\,m$$

C

$$120\,m$$

D

$$130\,m$$

Option B is Correct