Question

A particle is moving along X-direction. The formula of its displacement is x= 2t2 – 4t+9. Find out its velocity

and acceleration.

(A) (4t – 4), 4 (B) (2t – 4), 4 (C) t, 4 (D) t/2, 4​

Answer 1

Answer:-

The Correct Answer Is Option A) (4t - 4), 4.

Explanation:-

Given:-

• Displacement of the particle $\tt= 2t^2-4t+9.$

To Find:-

• Velocity and Acceleration of the particle.

IDs Used:-

$\\ \large\bf 1. \: \: \: \dfrac{d}{dx} x {}^{n} = nx {}^{n - 1}.$

$\large \bf2. \: \: \: \dfrac{d}{dx} y = 0. \: \: \tiny\textrm(if \: y \: is \: not \:depend \: on \: x).$

$\\ \large \bf 3. \: \: \: v = \frac{d}{dx} \: s. \\ \\ \tiny\bf where \: \: \begin{array}{|c c c|} \bf v & \bf = & \bf Velocity \\ \\ \bf s & \bf = & \bf Displacement \end{array} \:$

$\\ \large \bf 3. \: \: \: a = \frac{d}{dx} \: v. \\ \\ \tiny\bf where \: \: \begin{array}{|c c c|} \bf a & \bf = & \bf acceleration \\ \\ \bf v & \bf = & \bf velocity \end{array} \:$

So Here,

• s = 2t² - 4t + 9.

So Velocity,

$\\ \tt = \dfrac{d}{dt} \bigg(2{t}^{2} - 4t + 9 \bigg).$

$\\ \tt = 2 \: \frac{d}{dt} ( {t}^{2}) - 4 \: \frac{d}{dt} (t) + \frac{d}{dt} (9).$

$\\ \tt = 2 \times 2t {}^{2 - 1} - 4 \times 1t {}^{1 - 1} + 0.$

$\\ \tt = (2 \times 2t) - (4 \times 1)+ 0.$

$\\ \large\bf = 4t - 4 \: m/s.$

Therefore The Velocity is 4t - 4.

Similarly Acceleration,

$\\ \tt = \frac{d}{dx} v.$

$\\ \tt = \frac{d}{dt} \bigg(4t - 4 \bigg).$

$\\ \tt = \frac{d}{dt} (4t) - \frac{d}{dt} (4).$

$\\ \tt = 4 \times 1t {}^{1 - 1} + 0 .$

$\\ \large \bf = 4m / {s}^{2} .$

Therefore The Velocity And Acceleration Of The Particle Is 4t - 4 m/s and 4 m/s² Respectively.