Question

a student while doing
experiment finds that the velocity of an object
varies with time and it can be expressed as equation

v = x t² + yt + z.
if units V and t
are expressed in
terms of SI units, determine the
units of constants x,y and z in the
given
equation.​

Answer 1

Explanation:

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Answer 2

The S.I. units of $x, y$ and $z$ are $m/s^{3}, m/s^{2}$ and $m/s$ respectively.

Given,

Equation:$v=xt^{2} +yt+z$

v and t are in their S.I. units.

To find,

the units of constants x,y and z.

Solution:

• The principle that will be used here is known as "Principle of homogeneity of dimensions".
• It states that the dimensions of all the terms in an equation must be equal.
• Simple, it states that we add or subtract similar physical quantities.
• The dimensions of velocity, v can be found as:

$\frac{Distance}{Time} =\frac{[L]}{[T]} \\=[LT^{-1} ].$

The dimension of time is [T].

According to the principle of homogeneity,

Dimensions of $xt^{2}$=Dimensions of v

Dimensions of $yt$=Dimensions of v

Dimensions of $z$=Dimensions of v.

The dimensions of $x$ are,

$xt^{2} =v\\x[T^{2} ]=[LT^{-1} ]\\x=\frac{LT^{-1} }{T^{2} } \\x=[LT^{-3} ].$

As the S.I. units of velocity is $m/s$.

The S.I. units of $x$ will be $m/s^{3}$.

The dimensions of $y$ are,

$yt=v\\y[T]=[LT^{-1} ]\\y=\frac{[LT^{-1} ]}{[T]}\\ y=[LT^{-2} ].$

The S.I. units of $y$ will be $m/s^{2}$ (same as that of acceleration).

The dimensions of $z$ are,

$z=v\\z=[LT^{-1} ].$

The S.I. units of $z$ will be  $m/s$.

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