Question

The x and y coordinates of the particle at any time are x = 5t - 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is

Answer 1

Answer:

The acceleration of the particle at  t = 2s is -4 m/s².

Explanation:

Given that,

The x and y coordinates of the particle at any time.

$x=5t-2t^2$...(I)

$y=10t$...(II)

Acceleration :

The acceleration is the second derivative of the position of the particle.

Velocity :

The velocity is the firs derivative of the position of the particle.

We need to calculate the velocity of the particle

On differentiating w.r.to t of equation (I)

$v_{x}=\dfrac{dx}{dt}=5-4t$

Again differentiating

$a_{x}=\dfrac{d^2x}{dt}=-4$

We need to calculate the velocity of the particle

On differentiating w.r.to t of equation (II)

$v_{x}=\dfrac{dx}{dt}=10$

Again differentiating

$a_{x}=\dfrac{d^2x}{dt}=0$

Hence, The acceleration of the particle at  t = 2s is -4 m/s².

Answer 2

Answer:

the acceleration of the particle at t=2s is

$4m {s}^{ - 2}$

Explanation:

Given,

The x and y co-ordinates of the particle at any time are

$x = 5t - 2 {t}^{2} ........(i)$

$y = 10t.......(ii)$

Acceleration is second derivative of the position of the particle.

Velocity id the first derivative of the position of the particle.

We differentiating With respect to t equation (i)

$Vₓ = \frac{dx}{dt} = 5 - 4t$

Again differentiating,

$Aₓ = \frac{ {d}^{2} x}{d{t}^{2} } = - 4$

Again we differentiating equation (ii) with respect to t,

$Vₓ = \frac{dx}{dt} = 10$

And

$Aₓ = \frac{ {d}^{2} x}{d{t}^{2} } = 0$

Therefore the acceleration of the particle at t=2s is

$4m {s}^{ - 2}$