Question

if velocity v=Kx, then find acceleration at time t= 2 seconds​

Answer 1

$\huge\mathfrak\red{answer}$

It is given that

a= dtdv= dtd

(2t 2 e −t )=2[t 2

e −1(−1)+e −1 2t]

=e −1(4t−2t 2 )

So, a=0

⟹ 4t−2t 2 =0

Or 2t(2−t)=0

⟹ t=0 and t=2

Answer 2

The acceleration at time t= 2 seconds​ will be 0.

Given,

Equation:velocity v=Kx.

To find,

the acceleration at time t= 2 seconds​.

Solution:

• Velocity is equal to the rate of displacement of an object or the rate of change of positions.
• It is represented by the following expression:
• $v=\frac{ds}{dt}$ or $v=\frac{dx}{dt}$.
• where, ds-displacement and dx-position.
• Acceleration of an object is equal to the rate of change of velocity.
• It is represented by the following expression:
• $a=\frac{dv}{dt}$ .

The acceleration of the given particle will be:

$a=\frac{dv}{dt}\\a=\frac{d(Kx)}{dt} \\a=K.$

As the value of x comes to be a constant, it staes that the acceleration is not varying with time.

Thus, the acceleration will be 0.

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