Question

Position of particle is given by r = 3ti + 2t^2j + 3k where the is in the second and r in metre. Find 1) v(t) 2) a(t) Where v(t) and a(t) are velocity and acceleration of the particles at time t

Answer 1

Solution

The position of the particle is defined by the relation:

$\sf{\hat{r} = 3t \hat{i} + 2t^{2} \hat{j}=3 \hat{k} }$

To find

• Velocity of the particle
• Acceleration of the particle

Differentiating r w.r.t to t,we get velocity of the particle :

$\huge{\leadsto \ \underline{\boxed{\sf{v = 3 \hat{i} + 4t \hat{j]}}}}}$

Again differentiating v w.r.t to t,we get acceleration :

$\huge{\leadsto \ \underline{\boxed{\sf{a = 4 \hat{j}\ ms^{-2}}}}}$

Answer 2

Answer:

$\large\bold\red{(1)\;\vec{v}(t)=(3\hat{i}+4t\hat{j})\:m{s}^{-1}}\\\\\large\bold\red{(2)\;\vec{a}(t)=4\hat{j}\:m{s}^{-2}}$

Explanation:

Given,

position of a particle is given as,

$\large \boxed{\bold{\vec{r}(t) = 3t \hat{i} + 2 {t}^{2} \hat{j} + 3\hat{k}}}$

Where,

• r is in metres

And

• t is in seconds.

Now,

We have to find

• Velocity as function of time or v(t)

And

• Acceleration as function of time or a(t)

But,

We know that,

• Velocity is the derivative of position .

Therefore,

Differentiating the given Equation wrt time,

We get,

$= > \frac{d}{dt} \vec{r} = \frac{d}{dt}( 3t \hat{i} + 2 {t}^{2} \hat{j} + 3\hat{k}) \\ \\ = > \large \boxed{ \bold{\vec{v} (t)= (3\hat{i} + 4t\hat{j})\:m{s}^{-1}}}$

Further differentiating wrt time,

We get,

$= > \frac{d}{dt} \vec{v} = \frac{d}{dt} (3\hat{i} + 4t\hat{j})$

But,

We know that,

• Derivative of velocity is acceleration.

Therefore,

We get,

$= > \large \boxed{ \bold{\vec{a}(t) = 4\hat{j}\:m{s}^{-2}}}$