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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 \(=300+200=500\,m\)

 

Illustration Questions

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, \((x_f)=x_3\)

image

Change in position vector

\(\Delta x=x_f-x_i\)

\(=x_3-x_7\)

\(=-20-20\)

\(=-40\,m\)

image

For distance

\(s=\) total path covered

\(s=|x_9-x_7|+|x_3-x_9|\)

\(=|40-20|+|-20-40|\)

\(=20+60\)

\(=80\,m\)

image

Representation of the process

image image

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 \(x_3\).

image
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

Illustration Questions

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, \(x_f=x_{10}\)

image

 \(\Delta x=x_f-x_i\)

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

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

\(=70\,m\)

image

Average velocity \(=\dfrac{\Delta x}{\Delta t}\)

as \(\Delta t=20\,sec\)

\(\Delta x=70\,m\)

\(=\dfrac{70}{20}\)

\(=3.5\,m/sec\)

image

For distance \((s)\)

\(s\)= Total path covered

\(s=|x_{10}-x_3|\)

\(=|40-(-30)|\)

\(=70\,\,m\)

image

Average speed \(=\dfrac{s}{\Delta t}\)

\(=\dfrac{70}{20}\)

\(=3.5\,m/sec\)

image

\(\because\) Average velocity = Average speed

So, option (C) is correct.

image

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?

image
A

Average speed

.

B

Average velocity

C

Same

D

Can not be determined

Option C is Correct

Illustration Questions

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

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 :  \(|\Delta x|=|\Delta x_1|+|\Delta x_2|\)

\(\therefore\;s=|\Delta x|\)

Illustration Questions

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

Illustration Questions

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

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

Illustration Questions

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

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