Question

The distance through which a body falls from rest varies as square of time it takes to fall that distance. It is known that the body falls $64cm$ in $2$ seconds. How far does that body fall in $6$ seconds?

Answer 1

The statement " $x$ varies directly as $y$ ," means that when $y$ increases, $x$ increases by the same factor. In other words, $y$ and $x$ always have the same ratio that is:

$x=ky$

where $k$ is the constant of variation.

Let the distance be $d$ and the time be $t$.

Here, it is given that the distance $d$ varies as the square of time $t^2$ if $d=64$ cm and $t=2$ s, therefore,

$d=k{t}^2\Rightarrow 64=k{\left(2\right)}^2\Rightarrow 64=k\times 4\Rightarrow k=\frac{64}{4}\Rightarrow k=16$

Thus, the general equation is $d=16t^2$.

Since $t=6$ s, therefore,

$d=16{\left(6\right)}^2=16\times 36=576$

Hence, the distance that the body falls in $6$ seconds is $576$ cm.