Question

If displacement of the particle is proportional to the
third power of time i.e. S directly proportional to t^3 then its velocity will
be proportional to​

Answer 1

Explanation:

s=k×(t^3), where k is constant

v=ds/dt

v=3k(t^2)

then velocity proportional to t^2

Answer 2

Answer:

according to the question -

$s \: \alpha \: {t}^{3} \\( we \: know \: that \: derivative \: of \: s \: is \: equal \\ to \: velocity) \: . \\ there \: for \: \: \frac{ds}{dt} \: \alpha \: \frac{d {t}^{3} }{dt} \\ = > v \: \alpha \: \frac{dt}{dt} d {t}^{2} \\ = > v \: \alpha \: {t}^{2}$

so, velocity, v is directly proportional to the square of time, t.