Question

plz help me in this question ​

Answer 1

Step-by-step explanation:

x=t⁴/2-3t^2

v=dx/dt=2t³-6t

Velocity vanishes when v=0

or 2t³-6t=0

2t(t²-3)=0

t=0 or t=±√3

a=dv/dt=6t²-6

when t=0, a=-6m/s²

when t=±√3, a=12m/s²

Answer 2

Answer:

• At t = 0, a = -6 m/s²
• At t = ±√3, a = 12 m/s²

Step-by-step explanation:

Given :

Displacement is expressed as $\mathrm{\dfrac{t^4}{2} - 3t^2 }$

To find :

Acceleration at the time when the velocity vanishes

Solution :

Let the given expression of displacement be written as s(t)

$\longmapsto \mathrm{s(t) = \dfrac{t^4}{2} - 3t^2 }$

We know that,

$\boxed{\mathrm{\dfrac{ds}{dt} = v(velocity)}}$

So, differentiating, the given expression of displacement in terms of t, we get,

$\implies \mathrm{\dfrac{ds}{dt} = v = \dfrac{4t^3}{2} - 6t }$

$\because \boxed{\mathrm{\dfrac{d}{dx}x^n = nx^{n-1}}}$

So, we get,

$\implies \mathrm{v = 2t^3 - 6t}$

Now, we have to find the time where the velocity vanishes, that is  v = 0

$\implies \mathrm{v = 2t^3 - 6t =0}$

$\implies \mathrm{2t^3 - 6t = 0}$

$\implies \mathrm{2t(t^2 - 3) = 0}$

$\mathrm{Thus,\ t = 0\ (or) \ t = \pm \sqrt{3}}$

We need to find acceleration, at this point of time

We know that,

$\boxed{\mathrm{\dfrac{dv}{dt} = a(acceleration)}}$

So, differentiating, the given expression of velocity in terms of t, we get,

$\implies \mathrm{\dfrac{dv}{dt} =a= \dfrac{d}{dt}(2t^3 - 6t)}$

$\implies \mathrm{a = 6t^2-6}$

• At time, t = 0,

$\implies \mathrm{a = 6(0)^2-6}$

$\implies \underline{\boxed{\mathrm{a = -6\ m/s^2}}}$

• At time, t = ±√3

$\implies \mathrm{a = 6(\sqrt{3})^2-6}$

$\implies \mathrm{a = 6(3)-6}$

$\implies \mathrm{a = 18-6}$

$\implies \underline{\boxed{\mathrm{a = 12\ m/s^2}}}$

Hope it helps!!