Question

the position coordinates of an object are given by x =2t^2 and y=t^2-4t . the average velocity of particle in 1st one second ​

Answer 1

Given:

The position coordinates of an object are given by x =2t^2 and y=t^2-4t .

To find:

Average Velocity of particle in 1 second.

Calculation:

Average Velocity will be defined as the total Displacement divided by the time taken to cover the specific displacement.

In the x axis:

$x = 2 {t}^{2}$

$= > x_{0} = 2 \times {(0)}^{2} = 0 \: m$

$= > x_{1} = 2 \times {(1)}^{2} = 2 \: m$

So, Displacement

$\Delta x = x_{1} - x_{0}$

$= > \Delta x = 2- 0$

$= > \Delta x = 2 \: m$

Average Velocity in x axis:

$\boxed{ \bf{avg. \: v = \dfrac{\Delta x}{\Delta t} = \dfrac{2}{1} = 2 \: m {s}^{ - 1} }}$

In the y axis:

$y = {t}^{2} - 4t$

$= > y_{0} = {(0)}^{2} - 4(0) = 0 \: m$

$= > y_{1} = {(1)}^{2} - 4(1) = - 3 \: m$

So, Displacement

$\Delta y = y_{1} - y_{0}$

$= > \Delta y = - 3 - 0$

$= > \Delta y = - 3 \: m$

$\boxed{ \bf{avg. \: v = \dfrac{\Delta y}{\Delta t} = \dfrac{ - 3}{1} = - 3 \: m {s}^{ - 1} }}$

So, net average Velocity

$\therefore \: v = \sqrt{ {2}^{2} + {( - 3)}^{2} }$

$= > \: v = \sqrt{ 4 + 9}$

$= > \: v = \sqrt{ 13}$

So, final answer is:

$\boxed{ \sf{ \: v = \sqrt{ 13} \: m {s}^{ - 1} }}$

Answer 2

Question:-

The position coordinates of an object are given by x = 2t² and y = t² - 4t. Find the average velocity of particle in 1st one second

Answer:-

For x axis:-

Initial x-coordinate:- (at t = 0s)

x₁ = 2t²

=> x₁ = 2(0)²

=> x₁ = 0 m

Final x-coordinate:- (at t = 1s)

x₂ = 2t²

=> x₂ = 2(1)²

=> x₂ = 2 m

Displacement in x direction:-

Displacement = Δx

=> sₓ = x₂ - x₁

=> sₓ = (2 - 0)m

=> sₓ = 2m

Velocity in x direction:-

Velocity = (Δx) / (Δt)

=> vₓ = (sₓ) / (1 - 0)

=> vₓ = 2/1 m/s

=> vₓ = 2m/s

For y axis:-

Initial y-coordinate:- (at t = 0s)

y₁ = t² - 4t

=> y₁ = 0² - 4(0)

=> y₁ = 0 m

Final y-coordinate:- (at t = 1s)

y₂ = t² - 4t

=> y₂ = 1² - 4(1)

=> y₂ = -3 m

Displacement in y direction:-

Displacement = Δy

=> sᵧ = y₂ - y₁

=> sᵧ = (-3 - 0)m

=> sᵧ = -3m

Velocity in y direction:-

Velocity = (Δy) / (Δt)

=> vᵧ = (sᵧ) / (1 - 0)

=> vᵧ = (-3) / 1 m/s

=> vᵧ = -3 m/s

Net Velocity:-

vₙₑₜ = √[ (vₓ)² + (vᵧ)² ]

=> vₙₑₜ = √[ 2² + (-3)² ]

=> vₙₑₜ = √[ 4 + 9 ]

=> vₙₑₜ = √13 m/s

Ans. Average Velocity = √13 m/s