Question

A particle moves along x-axis in such a way that its coordinate x varies with time t according to the equation x=2-5t+6t2 .What is its initial velocity?

Answer 1

Particle is moving along x-axis so displacement is also along x-axis

Now , ⇒ x = 6t²-5t+2

since , V = dx/dt

Differentiating w.r.t. 't'

⇒ v = 6 × 2t - 5

⇒ v = 12t - 5

Answer 2

The initial velocity is -5 m/s

Given:

Equation of motion of a particle moving along x axis is given below:

$x = 2 - 5t + 6t^2$

Solution:

The calculation of initial velocity is done by differentiating the displacement with time.  

Therefore differentiating the equation,

$v=\frac{d x}{d t}$

$=\frac{d\left(2-5 t+6 t^{2}\right)}{d t}$

$v=-5+12 t$

Thereby,

For initial velocity, the time taken will be zero, thus, input value of time, t = 0

$v=-5+(12 \times 0)$

Initial velocity, v = -5 m/s