Question

The displacement x of a particle (in metre) is
related to time in second) by the relation
x = 2t - 37 + 2t + 2. Then velocity of particle
at the end of 2 second is
(A) 20 m/s
(B) 16 m/s
E) 14 m/s
(D) 8 m/s.​

Answer 1

Explanation:

b.

Prepare an essay on ENDANGERED ANIMALS and measures to be taken to

preserve them. You may note down appropriate information under these

headings

1) name of the animal.

10 places where they are found.

i interesting facts about the species.

iv) why do you think this species is endangered.

v) possible solutions to help this species, survive.

Answer 2

Correct Question :

The displacement x of a particle (in metre) is related to time (in second) by the relation

• x = 2t³ - 3t² + 2t + 2

Then the velocity of particle at the end of 2 second is...

Given :

Displacement equation of particle is given by

• x = 2t³ - 3t² + 2t + 2

To Find :

Velocity of particle at the end of 2 seconds.

Solution :

❖ In order to find velocity of the particle we have yo differentiate the given displacement equation with respect to time.

$\dag\:\underline{\boxed{\bf{\purple{v=\lim\limits_{t\to 0}^{}\:\dfrac{\Delta x}{\Delta t}=\dfrac{dx}{dt}}}}}$

$\sf:\implies\:v=\dfrac{dx}{dt}$

$\sf:\implies\:v=\dfrac{2t^3-3t^2+2t+2}{dt}$

$\sf:\implies\:v=\dfrac{d(2t^3)}{dt}-\dfrac{d(3t^2)}{dt}+\dfrac{d(2t)}{dt}+\dfrac{d(2)}{dt}$

$\sf:\implies\:v=6t^2-6t+2+0$

$\sf:\implies\:v=6t^2-6t+2$

Putting t = 2, we get .....

$\sf:\implies\:v=6(2)^2-6(2)+2$

$\sf:\implies\:v=6(4)-12+2$

$\sf:\implies\:v=24-10$

$:\implies\:\underline{\boxed{\bf{\orange{v=14\:ms^{-1}}}}}$

∴ (C) 14 m/s is the correct answer!