Question

will be marked as brainliest whoever answers correct

if
$\sqrt{x } = t - 2$
find displacement,instantaneous velocity and instantaneous acceleration​

Answer 1

Answer:

The displacement :

${t}^{2} - 4t + 4$

The instantanoeus velocity:

$(2t - 4) \frac{ \: length \: units}{timse \: units}$

And

the instanataneous acceleration is given by

$2\frac{ \: length \: units}{{(timse \: units )}^{2} }$

Explanation:

We have dispacement relation as ,

$\sqrt{x} = t - 2$

Where I'm assuming t is time and x is dispacement,

Now squaring both sides gives,

$x = {(t - 2)}^{2} \\ = {t}^{2} - 4t + 4 \: unit$

Hence the dispacement relation is gained.

Since instantanoeus velocity id defined as dispacement change in a small amount of time,

Taking the derivative of x will give us the instantaneous velocity,

$\frac{dx}{dt} = \frac{d {t}^{2} }{dt} - \frac{d(4t)}{dt} + \frac{d(4)}{dt} \\ \\ v_{ins} = 2t - 4 + 0 \\ \\ v_{ins} = 2t - 4$

Now you inst. Velcotiy has been gained.

And since instantaneous acceleration is defined as "rate if change of velocity in a really really small time" we gotta take the derivative of

v_{ins} to get the a_{ins}.

And we do that as,

(The units are always there I just have not written them for the sake of a clean looking calculation)

$a_{ins} = \frac{d(v_{ins}) }{dt} \\ \\ = \frac{d(2t - 4)}{dt} \\ \\ = \frac{d(2t)}{dt} - \frac{d(4)}{dt} \\ = 2 - 0 \\ = 2$

And your instantanoeus acceleration has been gained too.