##### Instantaneous Velocity and Acceleration
$\vec{v} = \lim_{\Delta t \to 0}\dfrac{\Delta s}{\Delta t}$ $\vec{a} = \lim_{\Delta t \to 0}\dfrac{\Delta v}{\Delta t}$

• To define and calculate the instantaneous velocity and speed.

• To differentiate between average and instantaneous velocity.

• To calculate the average acceleration and determine its direction.

• To calculate the instantaneous acceleration from position function or velocity function and determine its direction.

The instantaneous velocity is obtained as follows: $\vec{v} = \lim_{\Delta t \to 0}\dfrac{\Delta s}{\Delta t}$ Or, if s = f(t) then v = $\dfrac{d}{dt}f(t)$ = f’(t)
The instantaneous acceleration is obtained as follows: $\vec{a} = \lim_{\Delta t \to 0}\dfrac{\Delta v}{\Delta t}$ Or, if v = g(t) then a = $\dfrac{d}{dt}g(t)$ = $\dfrac{d^2}{dt^2}f(t)$

### Solutions for the topic Instantaneous Velocity and Acceleration

###### Solved Example: 28-1-01 Page 114

A particle moves along a straight line with the equation x = 16t + 4$t^2$ - 3$t^3$ where x is the distance in m and t is the time in second. Compute the acceleration of the particle after 2 seconds.

Solution:
\begin{align*} x &= 16t + 4t^2 - 3t^3\\ \dot{x} &= 16 + 4(2t) - 3(3t^2)\\ \ddot{x} &= 4(2) - 3(3)(2t) = 8 - 18t \end{align*} At t= 2 sec, $\ddot{x} = 8 - 18(2) = -28\ m/s^2$

###### Solved Example: 28-1-02 Page 114

A particle moves along a straight line such that distance (x) traversed in 't' seconds is given by $x= t^2*(t-4)$, the acceleration of the particle will be given by the equation:

Solution:
The acceleration can be found out by differentiating the displacement function twice with respect to time. $x= t^2*(t-4) = t^3 - 4t^2$ Differentiating, $v = 3t^2 - 8t$
Differentiating one more time, $a = 6t - 8$

###### Solved Example: 28-1-03 Page 114

The motion of a particle is defined by the relation x = (1/3)$t^3$ - 3$t^2$ + 8t + 2 where x is the distance in meters and is the time in seconds. What is the time when the velocity is zero?

Solution:
\begin{align*} x &= \dfrac{1}{3}t^3 - 3t^2 + 8t + 2\\ \dot{x} &= \dfrac{1}{3}\times 3 t^2 - 3\times 2t + 8 = t^2 - 6t + 8 = (t-4)(t-2) \end{align*} Hence, velocity is zero when t = 2 or t = 4

###### Solved Example: 28-1-04 Page 114

An equation x = $t^3$ - $2t$ denoted the relationship between displacement and time. At t = 4 sec. acceleration is given by:

Solution:
Velocity v= $\dfrac{dx}{dt}$ = $3t^2 -2$, acceleration a= $\dfrac{dv}{dt} = 6t$
At t = 4 sec, a = 24 units

###### Solved Example: 28-1-05 Page 114

The rate of change of displacement is called ________.

Solution:

The rate of change of displacement is velocity.

$\dfrac{ds}{dt} = v$

The rate of change of velocity is acceleration.

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

###### Solved Example: 28-1-06 Page 114

During uniform motion of an object along a straight line, its _______ remains constant with time.

Solution:
Since the motion is along a straight line, the direction of motion is NOT changing.
Also, since the motion is uniform, the distance convered in equal intervals is same too.
That means, the magnitude as well as the direction of the displacement is NOT changing, hence the velocity is constant.

###### Solved Example: 28-1-07 Page 114

If a body after travelling some distance comes back to its starting point then:

Solution:
Average velocity is given by: $\bar{v} = \dfrac{\Delta x}{\Delta t}$ Since $\Delta x = 0$ the average velocity is zero.

In scalar terms, Distance is related to speed.
In vector terms, displacement is related to velocity (and eventually to accelration)

Displacement considers only initial and final positions. Distance considers the path too.
Hence in this case, the displacement is zero, however distance traveled is NOT zero. So C is INCORRECT.

Since distance travelled in not zero, hence average speed is also NOT zero, hence option B is also INCORRECT.