Question

Displacement(y) of the particle is given by y=2t+t^2−2t^3 the velocity of the particle when acceleration is zero is given by *,​

Answer 1

Differentiation of y will give you velocity which is
2+2t-6t^2m/s

Then differentiate this to from MD acceleration which is
2-12tm/s^2

Then at acc = 0
-2 = -12t

T value comes out 1/6 sec
Put this in velocity

And velocity comes out to be 2/6 meters per second

Answer 2

Answer:

When acceleration is zero the velocity of the particle is  $\frac{13}{6} m /s$

Explanation:

Given: Displacement of a particle is given is $y = 2t + (t)^{2} - 2(t)^{3}$

To find: The velocity of a particle when acceleration is zero

Solution:

Displacement of a particle:

• The shortest distance between the two points  is referred to as displacement.
• It can be positive, negative, or zero because it is a vector quantity. Meters are the SI unit of displacement.
• It depends on the object's starting and ending positions; the starting point need not be the origin.

Displacement of a particle is given by

$y = 2t + (t)^{2} - 2(t)^{3}$

Differentiate the equation with respect to t

Velocity , v = $\frac{dy}{dt} = 2 + 2t - 6(t)^{2}$

Acceleration

a = $\frac{dv}{dt} = 2 - 12t$

Now acceleration is zero as per the given condition

a = 2 - 12t

sub a = 0

⇒ 0 = 2 - 12t

⇒ 12t = 2

⇒ $t = \frac{2}{12}$

⇒ $t = \frac{1}{6}$ sec

Putting this value in velocity equation

v = $2 + \frac{2}{6} - 6(\frac{1}{6} )^{2}$

v = $2 + \frac{1}{3} - \frac{1}{6}$

v = $\frac{13}{6} m /s$

Velocity of a particle v = $\frac{13}{6} m /s$

Final answer:

When acceleration is zero the velocity of the particle is  $\frac{13}{6} m /s$

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