Informative line

### Distance And Average Speed

Practice equation to calculate displacement & distance for one dimensional motion of the particle under constant & average acceleration. Learn formula for average speed & velocity problems equation.

# Displacement and Distance

## Displacement

• Displacement is defined as the change in the position vector of a particle. It is independent of path covered.
• Change in the position vector = Displacement
• Here in this diagram, Displacement $$=100-0$$ $$=100\,m$$

## Distance

• Distance is defined as the total path covered by a particle during the journey.
• So, total path covered by the particle as shown in figure $$=300+200=500\,m$$

#### Calculate displacement and distance for one dimensional motion of the particle, assuming that the particle starts moving from $$x_7$$, goes to $$x_9$$ and then returns to $$x_3$$.

A $$40\,m,\;80\,m$$

B $$-40\,m,\;80\,m$$

C $$-40\,m,\;-80\,m$$

D $$-20\,m,\;-70\,m$$

×

For displacement

Initial position of the particle, $$(x_i)=x_7$$

Final position of the particle, $$(x_f)=x_3$$

Change in position vector

$$\Delta x=x_f-x_i$$

$$=x_3-x_7$$

$$=-20-20$$

$$=-40\,m$$

For distance

$$s=$$ total path covered

$$s=|x_9-x_7|+|x_3-x_9|$$

$$=|40-20|+|-20-40|$$

$$=20+60$$

$$=80\,m$$

Representation of the process

### Calculate displacement and distance for one dimensional motion of the particle, assuming that the particle starts moving from $$x_7$$, goes to $$x_9$$ and then returns to $$x_3$$.

A

$$40\,m,\;80\,m$$

.

B

$$-40\,m,\;80\,m$$

C

$$-40\,m,\;-80\,m$$

D

$$-20\,m,\;-70\,m$$

Option B is Correct

# Average Velocity and Average Speed

Average velocity is defined as the ratio of total displacement to the total time taken by moving particle.

$$\text{Average velocity}=\dfrac{\Delta x}{\Delta t}$$     $$\big[\Delta x=\text{displacement},\;\Delta t=\text{total time taken}\big]$$

## Average Speed

Average speed is defined as the ratio of total path length to the total time taken by the moving particle to cover the distance.

$$\text{Average speed}=\dfrac{s}{\Delta t}$$

Note : Displacement and average velocity can either be positive or negative, whereas, both distance and speed, will always be positive.

$$\implies\;s\geq\Delta x$$

$$or, \,\;\text{Average speed}\geq\text{Average velocity}$$

• For one dimensional motion :

(i) If particle does not change its direction, then

Magnitude of average speed = Magnitude of average velocity

(ii) If particle changes its direction, then

Magnitude of average speed > Magnitude of average velocity

#### In the given diagram, if particle is moving in one dimensional motion and goes from $$x_3$$ to $$x_{10}$$ in $$20\,sec$$, then which one is more in magnitude?

A Average speed

B Average velocity

C Same

D Can not be determined

×

For displacement $$(\Delta x)$$

Initial position of particle, $$x_i= \;x_3$$

Final position of particle, $$x_f=x_{10}$$

$$\Delta x=x_f-x_i$$

$$=(x_{10}-x_3)\,$$

$$=[40-(-30)]\,\,$$

$$=70\,m$$

Average velocity $$=\dfrac{\Delta x}{\Delta t}$$

as $$\Delta t=20\,sec$$

$$\Delta x=70\,m$$

$$=\dfrac{70}{20}$$

$$=3.5\,m/sec$$

For distance $$(s)$$

$$s$$= Total path covered

$$s=|x_{10}-x_3|$$

$$=|40-(-30)|$$

$$=70\,\,m$$

Average speed $$=\dfrac{s}{\Delta t}$$

$$=\dfrac{70}{20}$$

$$=3.5\,m/sec$$

$$\because$$ Average velocity = Average speed

So, option (C) is correct.

### In the given diagram, if particle is moving in one dimensional motion and goes from $$x_3$$ to $$x_{10}$$ in $$20\,sec$$, then which one is more in magnitude?

A

Average speed

.

B

Average velocity

C

Same

D

Can not be determined

Option C is Correct

#### A ball is thrown vertically upward with a speed of $$100\,m/sec$$. What will be the average velocity of the ball after $$6\,sec$$?

A $$70\,m/sec$$

B $$100\,m/sec$$

C $$90\,m/sec$$

D $$75\,m/sec$$

×

Initial velocity $$v_0=100\;m/sec$$

Acceleration, $$a=-g$$

$$=-10\;m/sec^2$$

Displacement in $$6\,sec$$

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

$$=100×6+\dfrac{1}{2}\,(-10)×6^2$$

$$=600-(5×36)$$

$$=(600-180)\,$$

$$=420\,m$$

Average velocity $$=\dfrac{\Delta x}{t}\,$$

$$=\dfrac{420}{6}$$

$$=70\,m/sec$$

### A ball is thrown vertically upward with a speed of $$100\,m/sec$$. What will be the average velocity of the ball after $$6\,sec$$?

A

$$70\,m/sec$$

.

B

$$100\,m/sec$$

C

$$90\,m/sec$$

D

$$75\,m/sec$$

Option A is Correct

#### A ball is thrown vertically upward with a speed of 100 m/sec. Calculate the average speed of the ball after 12 sec.

A 52 m/sec

B 43.33  m/sec

C 50 m/sec

D 80.67 m/sec

×

Assuming upward motion always positive.

Initial velocity, u = 100 m/sec

a = – g = – 10 m/sec2

The direction of initial velocity and acceleration is opposite.

using, v = u+at

when v = 0

u = – at

$$t = \dfrac{u}{– a}$$

$$= \dfrac{100}{10}$$

= 10 sec

Displacement between 0 – 10 sec

using  $$v^2 = u^2 + 2as$$

$$\Delta x_1 = \dfrac{u^2}{- 2a}$$

$$= \dfrac{100 × 100}{2 × 10}$$

$$= 500 \; m$$

Displacement between 10 – 12 sec

$$\Delta x_ 2 = ut \; + \; \dfrac{1}{2} \; at^2 \;\\ at\; t=10\;sec, \; u = 0$$

$$\Delta x_2 = - \dfrac{1}{2} × 10 × (2)^2\\ = - 20 \; m$$

Total distance $$= | \Delta x_1 | + | \Delta x_2 |$$

$$= 500 + 20$$

$$= 520 \; m$$

$$\text{Average speed} = \dfrac{\text{Total distance }}{\text{Total time }}$$

$$= \dfrac{520}{12}$$

$$= 43 .33 \; m/sec$$

### A ball is thrown vertically upward with a speed of 100 m/sec. Calculate the average speed of the ball after 12 sec.

A

52 m/sec

.

B

43.33  m/sec

C

50 m/sec

D

80.67 m/sec

Option B is Correct

# Calculation of Distance

• If direction of the velocity does not change for one dimensional motion, then

Distance = Displacement

• If direction of the velocity changes, then initial velocity and acceleration will be in opposite direction.
• To calculate distance in this situation, follow the given steps :

Step-1 : Find magnitude of displacement before the velocity becomes zero i.e., $$|\Delta x_1|$$

Step-2 : Find magnitude of displacement after the velocity becomes zero i.e., $$|\Delta x_2|$$

Step-3 :  $$|\Delta x|=|\Delta x_1|+|\Delta x_2|$$

$$\therefore\;s=|\Delta x|$$

#### A particle is moving along a straight line with initial velocity $$20\,m/sec$$ and is subjected to an acceleration of $$-5\,m/sec^2$$. Calculate the distance covered by the particle in $$7\,sec$$.

A $$32.5\,m$$

B $$62.5\,m$$

C $$70.6\,m$$

D $$80\,m$$

×

Initial velocity, $$u=20\;m/sec$$

Acceleration, $$a=-5\;m/sec^2$$

The direction of acceleration and initial velocity is opposite.

Using, $$v=u+a\,t$$

when $$v=0$$

$$u=-a\,t$$

$$t=\dfrac{u}{-a}$$

$$=\dfrac{20}{5}$$

$$=4\,sec$$

Displacement between $$0-4\,sec$$

Using,  $$v^2=u^2+2\,as$$

$$\Delta x_1=\dfrac{(20)^2}{2×5}$$

$$=40\,m$$

Displacement between $$4-7\,sec$$

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

at $$t=4\,sec,\;u=0$$

$$\Delta x_2=\dfrac{-1}{2}×5×3^2$$

$$=-22.5\,m$$

Distance $$=|\Delta x_1|+|\Delta x_2|$$

$$=40+|-22.5|$$

$$=62.5\,m$$

### A particle is moving along a straight line with initial velocity $$20\,m/sec$$ and is subjected to an acceleration of $$-5\,m/sec^2$$. Calculate the distance covered by the particle in $$7\,sec$$.

A

$$32.5\,m$$

.

B

$$62.5\,m$$

C

$$70.6\,m$$

D

$$80\,m$$

Option B is Correct

# Motion of the Particle under Constant Acceleration

• If initial velocity = 0 and acceleration of particle is constant and particle does not change its direction then,

Magnitude of Displacement = Magnitude of Distance

#### A particle starting from rest has a constant acceleration of – 5 m/sec2. Calculate distance covered by the particle in 2 sec.

A 10 m

B – 10 m

C 20 m

D 30 m

×

Initial velocity, u = 0

Acceleration, a = – 5 m/sec2

Displacement $$(\Delta x)$$

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

Since, u = 0

$$\Delta x = - \dfrac{1}{2} × 5 × 2^2$$

$$= - 10 \; m$$

Distance $$(s)$$

$$s=| \Delta x|$$

$$= 10 \; m$$

Because, particle does not charge its direction.

### A particle starting from rest has a constant acceleration of – 5 m/sec2. Calculate distance covered by the particle in 2 sec.

A

10 m

.

B

– 10 m

C

20 m

D

30 m

Option A is Correct