Question

Moving body covers a distance proportional to cube of its time period the accelaration of body is

Answer 1

Heyy!!
Here is your answer

Let the distance covered in time 't' be 'x' units

According to the question,
x proportional to t^3
=>
$x = k \: {t}^{3} \\where \: k \: is \: proportionity \: \\ constant \: \\ \frac{dx}{dt} \: = k \: \times 2 {t}^{2} \\ = > v \: \: = 2k {t}^{2} \\ now \: for \: acceleration \\ a \: = \frac{dv}{dt} = 2k \times 2t \\ a = 4kt$
a α t
Now finally we can say that here
distance covered will be directly proportional to time peroid 't'.

Hope you understand my answer and it may helps you!