Question

A particle moving along a straight line with constant acceleration of -2m/s² has a velocity of 6m/s at time = 0s ,find the ratio of distance and displacement at time=5s.
Please give full solution in paper.​

Answer 1

The ratio of the distance and displacement of the particle in t = 5 seconds is equal to 5 : 13.

Given:

The initial velocity of the particle (u) = 6 m/s

The acceleration of the particle (a) = -2 ms⁻²

Total time (T) = 5 seconds

To Find:

The ratio of distance and displacement in 5 seconds.

Solution:

(1.) Calculating the displacement of the particle in t = 5 seconds:

The initial velocity of the particle (u) = 6 m/s

The acceleration of the particle (a) = -2 ms⁻²

Total time (T) = 5 seconds

→ Using the second equation of motion:

                                     $\boldsymbol{S=ut+\frac{1}{2}at^{2}}$

                                 $\boldsymbol{\therefore S=(6)(5)+\frac{1}{2} \times(-2)\times(5^{2} )}\\\\\boldsymbol{\therefore S=30-25}\\\\\boldsymbol{\therefore S=5\:m}$

→ The displacement of the particle in t = 5 seconds is equal to 5 m.

(2.) Calculating the distance travelled by the particle in t = 5 seconds:

• Since the acceleration of the particle is negative, therefore the speed of the particle is continuously decreasing and it will become zero at some point in time.
• The speed of the particle, after reaching the minimum value will once again start increasing in the reverse direction and thus for this time interval, the particle will travel in the reverse direction.

Let the speed of the particle becomes zero after time 't'.

→ Using the second equation of motion:

                                        $\boldsymbol{v=u+at}$

                                     $\boldsymbol{\therefore 0=6+(-2)(t)}\\\\\boldsymbol{\therefore t=\frac{6}{2} =3\:seconds}$

Therefore the speed of the particle becomes zero after 3 seconds.

(i) Calculating the distance travelled by the particle in t = 3 seconds:

                                         $\boldsymbol{S=ut+\frac{1}{2}at^{2}}$

                                     $\boldsymbol{\therefore S=(6)(3)+\frac{1}{2} \times(-2)\times(3^{2} )}\\\\\boldsymbol{\therefore S=18-9}\\\\\boldsymbol{\therefore S=9\:m}$

∴ The particle travels a distance of 9 m in the first 3 seconds.

→ The speed of the particle becomes zero after 3 seconds.

(ii) Calculating the distance travelled by the particle in the next 2 seconds:

                                         $\boldsymbol{S=ut+\frac{1}{2}at^{2}}$

                                     $\boldsymbol{\therefore S=(0)(2)+\frac{1}{2} \times(-2)\times(2^{2} )}\\\\\boldsymbol{\therefore S=0-4}\\\\\boldsymbol{\therefore S=-4\:m}$

∵ The displacement of the particle is equal to -4 m.

∴ The distance travelled by the particle in the next 2 seconds = 4 m

∴ The distance travelled by the particle in 5 seconds = 9 + 4 = 13 m

∴ The ratio of distance and displacement in 5 seconds = 5/13

Therefore the ratio of the distance and displacement in t = 5 seconds is equal to 5 : 13.

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