Physics, asked by Barbossa19, 10 months ago

A Particle moves along a straight line such that its displacement at any time t is given by
s = (t^3 - 6t^2 + 3t +4) m
The velocity when the acceleration is zero. ​

Answers

Answered by Anonymous
87

Given :

▪ Displacement-time equation of a particle has been provided.

\bigstar\:\underline{\boxed{\bf{\green{s=t^3-6t^2+3t+4}}}}

To Find :

▪ Velocity of particle when the acceleration is zero.

SoluTion :

Given,

\dashrightarrow\sf\:s=t^3-6t^2+3t+4

Therefore,

\dashrightarrow\sf\:velocity(v)=\dfrac{ds}{dt}=3t^2-12t+3

Acceleration a is given by

\dashrightarrow\sf\:a=\dfrac{dv}{dt}\\ \\ \dashrightarrow\sf\:a=6t-12\\ \\ \dashrightarrow\sf\:0=6t-12\\ \\ \dashrightarrow\sf\:\red{t=2\:s}

Hence, at t = 2 s the velocity will be

\dashrightarrow\sf\:v=3t^2-12t+3\\ \\ \dashrightarrow\sf\:v=3(2)^2-12(2)+3\\ \\ \dashrightarrow\sf\:v=12-24+3\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{v=-9\:ms^{-1}}}}}\:\orange{\bigstar}

Answered by BrainlySmile
68

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s =  \frac{1}{2} at² + ut

3) v² - u² = 2as

6. Average Speed = Total distance ÷ Total time

7. Average Velocity = Total displacement ÷ Total time

8. Some basic formulae using calculus method:

1) Velocity =  \frac{dx}{dt}

2) Acceleration =  \frac{dv}{dt}

3) Acceleration (when x is given) =  v \frac{dv}{dx}

Given question: A particle moves along a straight line such that its displacement at any time t is given by s = (t³ - 6t² + 3t + 4) m.

Fine the velocity when the acceleration is zero. ​

Answer: s = (t³ - 6t² + 3t + 4) m

We know that v =   \frac{ds}{dt}

On differentiating with respect to t, we get,

v = (3t² - 12t + 3) m/s

We also know that, a =  \frac{dv}{dt} .

Again on differentiating with respect to t, we get,

a = (6t - 12) m/s²

It is given that a = 0 m/s²

⇒ 0 = 6t - 12

⇒ 6t = 12

⇒ t = 2 seconds

So, substituting the value of t in v, we get,

v = [3(2)² - 12(2) + 3] m/s

v = [12 - 24 + 3] m/s

v = - 9 m/s

∴ Velocity will be - 9 m/s when acceleration = 0 m/s².

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