Notes - Distance and Displacement
Category : NEET
Distance and Displacement
1) Distance: It is the actual path length covered by a moving particle in a given interval of time.
(i) If a particle starts from A and reach to C through point B as shown in the figure.
Then distance travelled by particle \[=AB+BC=7\,m\]
(ii) Distance is a scalar quantity.
(iii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iv) Unit: metre (S.I.)
(2) Displacement: Displacement is the change in position vector i.e., A vector joining initial to final position.
(i) Displacement is a vector quantity
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iii) Unit: metre (S.I.)
(iv) In the above figure the displacement of the particle \[\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\]
\[\Rightarrow \,\,\left| AC \right|=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}+2(AB)\,(BC)\,\cos \,{{90}^{o}}\,}=\,\,5\,m\]
(v) If \[{{\vec{S}}_{1}},\,{{\vec{S}}_{2}},\,{{\vec{S}}_{3}}\,........\,{{\vec{S}}_{n}}\] are the displacements of a body then the total (net) displacement is the vector sum of the individuals. \[\vec{S}={{\vec{S}}_{1}}+\,{{\vec{S}}_{2}}+\,{{\vec{S}}_{3}}+\,........\,+{{\vec{S}}_{n}}\]
(3) Comparison between distance and displacement:
(i) The magnitude of displacement is equal to minimum possible distance between two positions.
So \[distance\ge \left| Displacement \right|\].
(ii) For a moving particle distance can never be negative or zero while displacement can be.
(zero displacement means that body after motion has came back to initial position)
i.e., Distance > 0 but Displacement > = or < 0
(iii) For motion between two points displacement is single valued while distance depends on actual path and so can have many values.
(iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position.
(v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction.
(vi) If \[{{\vec{r}}_{A}}\,\,and\,\,{{\vec{r}}_{B}}\] are the position vectors of particle initially and finally.
Then displacement of the particle
\[{{\vec{r}}_{AB}}={{\vec{r}}_{B}}-{{\vec{r}}_{A}}\]
and s is the distance travelled if the particle has gone through the path APB.
Sample problems based on distance and displacement
Problem 1. A man goes 10 m towards North, then 20m towards east then displacement is
[KCET (Med.) 1999; JIPMER 1999; AFMC 2003]
(a) 22.5 m (b) 25 m (c) 25.5 m (d) 30 m
Solution: (a) If we take east as \[x-\] axis and north as \[y-\] axis, then displacement \[=20\,\hat{i}\,+\,10\hat{j}\]
So, magnitude of displacement \[=\sqrt{{{20}^{2}}+{{10}^{2}}}=10\sqrt{5}=\text{ }22.5\,\,m.\]
Problem 2. A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be respectively
(a) \[\frac{\pi r}{2},\,r\sqrt{2}\] (b) \[\frac{\pi r}{4},\,r\] (c) \[\pi r,\,\frac{r}{\sqrt{2}}\] (d) \[\pi r,\,r\]
Solution: (a) Let particle start from A, its position vector \[{{\vec{r}}_{O}}_{A}=r\hat{i}\]
After one quarter position vector \[{{\vec{r}}_{OB}}=r\,\hat{j}.\]
So displacement \[=r\hat{j}-r\hat{i}\]
Magnitude of displacement \[=r\sqrt{2}.\]
and distance = one fourth of circumference \[=\frac{2\pi r}{4}=\frac{\pi r}{2}\]
Problem 3. The displacement of the point of the wheel initially in contact with the ground, when the wheel roles forward half a revolution will be (radius of the wheel is R)
(a) \[\frac{R}{\sqrt{{{\pi }^{2}}+4}}\] (b) \[R\sqrt{{{\pi }^{2}}+4}\] (c) \[2\pi R\] (d) \[\pi R\]
Solution: (b) Horizontal distance covered by the wheel in half revolution = pR
So the displacement of the point which was initially in contact with a ground =\[\sqrt{{{(\pi R)}^{2}}+{{(2R)}^{2}}}\]
\[=R\sqrt{{{\pi }^{2}}+4}.\]
You need to login to perform this action.
You will be redirected in
3 sec
