Question

1. The motion of a particle is described by
$x = xo(1 - {e}^{ - kt} ) \: where \: t \geqslant 0 \: \: \: \: \: \: \: \: xo > 0 \: \: \: \: \: \: \: \: \: k > 0$
With what
velocity does the particle start?
$(a) \frac{xo}{k} \\ (b) xok\\( c) \frac{k}{xo} \\ ( d)2xok$

​

Answer 1

Given:

The motion of a particle is described by

$\sf\:x=x_o(1-e^{-kt})$ where $\sf\:t\geqslant0\:and\:x>0\:and\:k>0$

To Find :

Velocity of the particle at t = 0 sec

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

$\sf\:Velocity=\dfrac{Distance}{Time\:interval}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

Solution :

We have ,

$\sf\:x=x_o(1-e^{-kt})$

Now , Differentiate it with respect to t

$\sf\implies\dfrac{dx}{dt}=x_o[\dfrac{d(1)}{dt}-\dfrac{e^{-kt}}{(-kt)}\times\dfrac{d(-kt)}{dt}]$

$\sf\implies\dfrac{dx}{dt}=x_o[0-e^{-kt}\times(-k)]$

$\sf\implies\dfrac{dx}{dt}=x_o\:k\:e^{-kt}$

$\sf\implies\dfrac{dx}{dt}=x_o\:k\times\dfrac{1}{e^{kt}}$

$\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{kt}}$

At t = 0

$\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{k\times0}}$

$\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{0}}$

$\sf\implies\:v=x_o\:k$

Therefore , correct option b )

__________________

More information about topic

• Velocity is a vector quantity.
• The velocity of an object can be positive, zero and negative.
• SI unit of Velocity is m/s
• Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$

Answer 2

Answer:

Given:

The motion of a particle is described by

\sf\:x=x_o(1-e^{-kt})x=x

o

(1−e

−kt

) where \sf\:t\geqslant0\:and\:x > 0\:and\:k > 0t⩾0andx>0andk>0

To Find :

Velocity of the particle at t = 0 sec

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

\sf\:Velocity=\dfrac{Distance}{Time\:interval}Velocity=

Timeinterval

Distance

In differential form:

\sf\:Velocity,V=\dfrac{dx}{dt}Velocity,V=

dt

dx

Solution :

We have ,

\sf\:x=x_o(1-e^{-kt})x=x

o

(1−e

−kt

)

Now , Differentiate it with respect to t

\sf\implies\dfrac{dx}{dt}=x_o[\dfrac{d(1)}{dt}-\dfrac{e^{-kt}}{(-kt)}\times\dfrac{d(-kt)}{dt}]⟹

dt

dx

=x

o

[

dt

d(1)

−

(−kt)

e

−kt

×

dt

d(−kt)

]

\sf\implies\dfrac{dx}{dt}=x_o[0-e^{-kt}\times(-k)]⟹

dt

dx

=x

o

[0−e

−kt

×(−k)]

\sf\implies\dfrac{dx}{dt}=x_o\:k\:e^{-kt}⟹

dt

dx

=x

o

ke

−kt

\sf\implies\dfrac{dx}{dt}=x_o\:k\times\dfrac{1}{e^{kt}}⟹

dt

dx

=x

o

k×

e

kt

1

\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{kt}}⟹v=x

o

k×

e

kt

1

At t = 0

\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{k\times0}}⟹v=x

o

k×

e

k×0

1

\sf\implies\:v=x_o\:k\times\dfrac{1}{e^{0}}⟹v=x

o

k×

e

0

1

\sf\implies\:v=x_o\:k⟹v=x

o

k

Therefore , correct option b )

__________________

More information about topic

Velocity is a vector quantity.

The velocity of an object can be positive, zero and negative.

SI unit of Velocity is m/s

Dimension of Velocity: \sf\:[M^0LT{}^{-1}][M

0

LT

−1

]