Question

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The Displacement of a Particle
$x = 4t {}^{2} - 15t + 25$
Find the Position, Velocity and acceleration of particle at T=0s.

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Answer 1

Given :-

Displacement of particle is

$x = 4t^2 -15 t + 25$

To find :-

Position, Velocity and acceleration.

Solution:-

Position at t = 0 s

→$x = 4t^2 -15 t +25$

Put t = 0

→$x = 4 (0)^2 -15 \times 0 +25$

→$x = 25 m$

hence,

The position of the particle will be 25 m.

Now,

$\huge \boxed{v = \dfrac{dx}{dt}}$

→$v = \dfrac{d(4t^2 -15t +25)}{dt}$

→$v = 4\times 2t -15 +0$

→$v = 8t -15$

Thus , at t = 0 s

Velocity will be

→$v = 8 \times 0 -15$

→$v = -15 m/s$

Now,

Acceleration is given by :-

$\huge \boxed{a = \dfrac{dv}{dt}}$

→$a = \dfrac{d(8t -15) }{dt}$

→$a = 8 \times 1 -0$

→$a = 8 m/s^2$

at t = 0

the acceleration will be 8 m/s².

Answer 2

SOLUTION

Given,

$= > x = 4 {t}^{2} - 15t + 25$

Putting the value of t= 0 in this equation.

$= > x = 4(0) {}^{2} - 15(0) + 25 \\ = > x = 0 - 0 + 25 \\ = > x = 25m$

So, position, x= 25m

Velocity:

$= > \frac{dx}{dt} = 8t - 15 \\ = > t = 0 \: </u></strong><strong><u>,</u></strong><strong><u>v = 0 - 15 \\ = > v = - 15m</u></strong><strong><u>/</u></strong><strong><u>s</u></strong><strong><u>$

Acceleration:

$= > a = \frac{dx}{dt} = 8ms {}^{ - 2}$

Hope it helps ☺️