Question

Displacement of a particle is given by x=3 t^2+ 2 t. Determine the instantaneous velocity.Also find instantaneous acceleration.

Answer 1

Answer:

Displacement function of object is given as :

x = 3t² + 2t

We need to find out the :

• Instantaneous Velocity
• Instantaneous Acceleration

Calculation:

We get Instantaneous velocity by 1st order derivative of displacement function.

$x = 3 {t}^{2} + 2t$

$= > v = \dfrac{dx}{dt}$

$= > v = \dfrac{d(3 {t}^{2} + 2t) }{dt}$

$= > v = 6t + 2$

Now we shall get Instantaneous Acceleration by 1st order derivative of velocity or 2nd order derivative of displacement function.

$a = \dfrac{dv}{dt}$

$= > v = \dfrac{d(6t + 2)}{dt}$

$= > a = 6 \: units$

Answer 2

TOPIC :- Differential kinematics

Concept

In variable kinematics, acceleration is not constant .Hence, we can not apply basic laws of motion as they are derived under the constrains that either acceleration is zero or constant.Although it may be possible that acceleration is constant but there may be more constraints like displacement is not a linear function of velocity, velocity dependent on displacement etc.

We solve problems with variable acceleration using differential calculus using the fact that acceleration at infinitesimal time is constant.Here, we use some important formulas which are :-

$v = \frac{dx}{dt} \\ \\ a = \frac{dv}{dt} \\ \\ \\ a = v \frac{dv}{ds}$

SOLUTION :-

Given,

x = 3t² + 2t

Hence , v = $\frac{dx}{dt} </p><p></p><p>= \frac{d(3 {t}^{2} + 2t) }{dt} \\</p><p></p><p> = \frac{d(3 {t}^{2}) }{dt} + \frac{d(2t)}{dt } \\ </p><p></p><p>= 6t + 2$

Hence, instantaneous velocity is 6t + 2 .

And,

a = $\frac{dv}{dt} = \frac{d(6t + 2)}{dt} \\ = 6$

Hence, instantaneous acceleration is 6 unit