Question

Define instantaneous velocity. (b) the position of an object moving along x-axis is given by x=8.5+2.5t^2 ¡) what is its velocity at t=2.0s .

Answer 1

Iñstãntaneous Velocity

Velocity of a particle at certain iñstánce of time in the whole time interval is known as Iñstãntaneous Velocity

In other words,it is the limit of average velocity

Mathematically,it is represented by:

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

From the Question,

The position "x" of the particle is defined by relation:

$\sf{x = 8.5 + 2.5t {}^{2} }$

To find velocity of the particle at t = 2s

Differentiating x w.r.t to t,we get:

$\sf{v = \dfrac{dx}{dt} } \\ \\ \longrightarrow \ \: \sf{v = \dfrac{d(8.5 + 2.5t {}^{2}) }{dt} } \\ \\ \longrightarrow \: \sf{v = 5t}$

At t = 2,

$\implies \: \underline{ \boxed{ \sf{v = 10ms {}^{ - 1} }}}$

The velocity of the particle at t = 2s is 10m/s.

Answer 2

Solution:

A).

• Instantaneous Velocity: Velocity of a particle at a particular moment in time.

B). x = 8.5 + 2.5t²

$\sf{\implies Velocity=\dfrac{dx}{dt}}$

$\sf{\implies Velocity=\dfrac{d}{dt}(2.5t^{2}+8.5)}$

$\sf{\implies Velocity=\dfrac{d}{dt}(2.5t^{2})+\dfrac{d}{dt}(8.5)}$

$\star{\boxed{\sf{Now,\;by\;using\;identity=\dfrac{d}{dx}x^{n}=nx^{n-1}}}}$

$\sf{\implies Velocity=\dfrac{d}{dt}(2.5t^{2}+8.5)}$

$\sf{\implies Velocity=2(2.5^{2-1})+0}$

$\sf{\implies Velocity=5t}$

Now, at t = 2

$\sf{\implies (v)_{t=2}=5(2)}$

$\sf{\implies Velocity=10\;m/s}$

Hence, the velocity of the particle at t = 2s is 10 m/s