Test: Position, Path Length and Displacement- 2 - EmSAT Achieve MCQ

• When the object hasn't moved in a straight line from its initial to final position. The displacement being shorter than the traveled path results in a ratio less than one.
• When an object moves in a straight line without changing its direction. In such a scenario, the displacement equals the total distance.
• Therefore, the displacement can never be more than the distance, resulting in the ratio being equal to or less than one.

Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)

a² + b² = c²

Let AB be the displacement of the person towards the North, BC be the displacement towards the east and CE be the displacement in the South-West direction. We have to find the displacement of the person from the origin, EA

Consider the right triangle Δ ECD, by Pythagoras' theorem

ED² + CD² = EC² → (i)

From fig, CD=AB=30m

EC=30√2 m

ED=EA+AD

AD=BC=20m

∴ED=EA+20m

(i)→(EA+20)² + (30)² = (30√2)²

(EA+20)² + 900  = 1800

EA+20=30

EA=10m

The correct answer is option (3).

• Angular displacement (θ) is defined as total angle displaced by the particle from its starting position and it is given as radian or degrees 
• The unit of angular displacement is rad or Degree.
• Whereas Radian is a dimensionless number, so its angular displacement won’t have any dimension hence it is dimensionless quantity.

Remember that with angular displacement, counter-clockwise is positive and clockwise is negative conventionally.

Hence the correct option is positive for counter-clockwise.

When we drop the second ball till this time the first ball has some velocity, which is greater than the second ball,

 v > v’ where v is the velocity of second ball and v’ is the velocity of the first ball.

• v_net > 0. Where v_net is the net difference between both velocities.
• the motion is in the vertical direction and gravitational acceleration ‘g’ is acting same on both balls, but the first ball is dropped from a height, and its travel ‘s’ distance (suppose)  and acquired a velocity ‘v’ but,

The second ball released when the first ball at ‘s’ distance, then second has less velocity because it travels less distance than the first ball.

So, v_net > 0. And,

Both balls are moving under the same gravitational acceleration ‘g’,

∴ a_net = 0. Where a_net is the net difference of acceleration.

So, before hitting the grounds both balls have a displacement and it is increasing in nature. So, option 1 will be correct.

Distance:

• The measurement of how much any object moved is called distance.
• It is a scalar quantity.

Displacement:

• The minimum distance between two points is called the displacement of the object.
• It is a vector quantity.

Given:

Radius of semicircular track (r) = 140 m

• Displacement of the semicircular track is equal to the diameter of the circular track i.e. 

⇒ Displacement  = 2R = 2 × 140 = 280 m

Distance (S): 

• The total distance traveled by an object from a starting point is called path length. Thus the total path length of an object is called distance traveled by that object.
• Distance is equal to the magnitude of displacement sometimes.
• Distance is greater than equal to zero always. It can never be negative.
• It is a scalar quantity.
• The SI unit of distance is meter (m).

Displacement (S’):

• The minimum path length between starting point to final point is called displacement.

• Displacement is a vector quantity.
• It can be positive, negative, or zero.
• The SI unit of displacement is meter (m).

Calculation:

The Car took the semi-circular path of a semi-circle of diameter d = 50 m

The distance travelled is circumference C = π d = π (50 m )

The shortest distance is displacement which is the straight line between the final point and starting point. This the diameter of semi circular path.

diameter d = 50 m

So, displacement s = 50 m

Distance = π r = π × (50/2) = 25π 

The ratio is 

25π / 50 = π : 2

So, the correct option is  π : 2

Distance:

• The measurement of how much any object moved is called distance.
• It is a scalar quantity.

Displacement

• It is the shortest distance between the initial and final positions of the particle.
• It is a vector quantity.

Let the initial position of the boy is considered as the origin.

So the initial position vector of the boy is,

    -----(1)

So according to the diagram, the final position vector of the boy is given as,

     -----(2)

By equation 1 and 2 the resultant displacement of the boy is given as,

⇒ d = 13 m

• Hence, option 1 is correct.

CONCEPT:

• Displacement: Displacement is a vector quantity, which is the shortest distance between the initial and final points.
  • The unit is in M.K.S system is ‘meter’.
  • The dimensional formula is [L].

• Vector: it is the physical quantity that has direction as well as magnitude. For example – Force, weight, Electric field, etc.

CALCULATION:

Given that walk 8 m east then walk 6 m north.

So, we know that angle between north and east is 90°.

So, apply Pythagoras theorem:

d² = a² + b²

So,

d² = 8² + 6² = 64 + 36 = 100

d = 10 meter.

Displacement is velocity multiplied by time. 

Thus the area under the velocity-time graph gives the displacement. Similarly, the area under-speed time graph gives the total distance travelled

Given:

v(t) = (1 - 3t2 + 2t3) m/s

its position at t = 0 is x = 0 then at t = 2 s, its position is to be calculated.

Since, 

Now,