Distance and Displacement-
Before you go through this article, make sure that you have gone through the previous article on Distance and Displacement.
We have discussed-
- Distance is the total length of the actual path travelled by a particle during its motion.
- Displacement is a vector drawn from the initial position to the final position of the particle.
In this article, we will discuss problems on distance and displacement.
PRACTICE PROBLEMS BASED ON DISTANCE & DISPLACEMENT-
Problem-01:
A particle covers half of the circle of radius R. Then, the displacement and distance of the particle are respectively-
- 2πR, 0
- 2R, πR
- πR/2, 2R
- πR, R
Solution-
Consider particle moves from point A to point B covering half of the circle as shown-
Displacement-
Magnitude of displacement of the particle
= Shortest distance between point A and point B
= Diameter of the circle
= 2R units
Distance-
Distance covered by the particle
= Total length of the actual path
= Circumference of the circle / 2
= 2πR / 2
= πR units
Thus, Option (B) is correct.
Problem-02:
A particle moves in a quarter circle of radius R. What are the values of distance and displacement respectively?
- πR/2, √2R
- πR, R
- 2√2R, πR/2
- πR/2, 0
Solution-
Consider particle moves from point A to point B covering half of the circle as shown-
Distance-
Distance covered by the particle
= Total length of the actual path
= Length of arc AB
= π/2 x R { Using the relation θ = arc / radius }
= πR/2 units
Displacement-
Magnitude of displacement of the particle
= Shortest distance between point A and point B
= Length of line AB
= √(OA2 + OB2)
= √(R2 + R2)
= √2R units
Thus, Option (A) is correct.
Problem-03:
A mosquito flies from a point A(-1, 2, 3) m to a point B(2, -1, -2) m. Find the displacement of the mosquito.
Solution-
We have-
- Initial position vector,
- Final position vector,
Displacement vector of the mosquito is given by-
So, we have-
Thus, displacement vector of the mosquito is-
The magnitude of displacement is given by-
||
= √(32 + 32 + 52)
= √(9 + 9 + 25)
= √43 m
Thus, magnitude of displacement of the mosquito = √43 m.
Problem-04:
A mosquito flies from corner A to the corner B of a cube of side length 5 m moving along its sides as shown. Find the displacement of the mosquito.
Solution-
Consider the origin of Cartesian coordinate system to be present at point A.
Then-
- Coordinates of point A = (0, 0, 0)
- Coordinates of point B = (5, 5, 5)
Displacement vector of the mosquito is given by-
Thus, displacement vector of the mosquito is-
The magnitude of displacement is given by-
||
= √(52 + 52 + 52)
= √(3 x 52)
= 5√3 m
Thus, magnitude of displacement of the mosquito = 5√3 m.
Problem-05:
A particle is moving in a circular path of radius R. The distance and displacement of the particle when it describes an angle of 60° from its initial position respectively are-
- πR/3, R
- πR/6, √2R
- πR/3, √2R
- πR/3, √2R
Solution-
Consider particle moves from point A to point B covering an angle of 60º from its initial position as shown-
Distance-
Distance covered by the particle
= Total length of the actual path
= Length of arc AB
= π/3 x R { Using the relation θ = arc / radius }
= πR/3 units
Displacement-
In Δ AOB,
- OA = OB = R
- ∠AOB = 60°
Now,
- We know, angles opposite to equal sides are equal. So, ∠OAB = ∠OBA.
- Also, we know sum of three angles in a triangle is 180°.
- So, we conclude ∠OAB = ∠OBA = 60°.
Since all the angles in ΔAOB are 60°, so ΔAOB is an equilateral triangle.
So, all the sides length must be same.
Thus, length of line AB = R units.
Now,
Magnitude of displacement of the particle
= Shortest distance between point A and point B
= Length of line AB
= R units
Thus, Option (A) is correct.
Problem-06:
A player completes a circular path of radius R in 40 seconds. Its distance and displacement at the end of 2 minutes 20 seconds will be-
- 7πR, 2R
- 2R, 2R
- 2πR, 2R
- 7πR, R
Solution-
We have number of revolutions made by the player in 40 seconds = 1.
So, number of revolutions made by the player in 2 minutes 20 seconds (140 seconds)
= x 140
= 3.5 revolutions
Distance-
Distance covered in 3.5 revolutions
= 3.5 x Distance covered in 1 revolution
= 3.5 x Circumference of the circle
= 3.5 x 2πR
= 7πR units
Displacement-
Magnitude of displacement after 3.5 revolutions
= Diameter of the circle
= 2R units
Thus, Option (A) is correct.
Problem-07:
Given that P is a point on a wheel rolling on horizontal road. The radius of the wheel is R. Initially, the point P is in contact with ground. The wheel rolls through half of the revolution. What is the displacement of point P?
Solution-
Given-
- Initially, the point P is in contact with ground.
- The wheel rolls through half of the revolution.
After rolling through half revolution, point P is at the top most point of the wheel.
So, we have-
Here, d represents the displacement of point P.
Using Pythagoras theorem, we have-
d2 = (2R)2 + (πR)2
d2 = 4R2 + π2R2
d2 = (4+π2)R2
∴ d = R√(4+π2)
Thus, magnitude of displacement of point P = R√(4+π2) m.
To practice more problems on distance and displacement,
Next Article- Speed and Velocity
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