Acceleration – Average and Instantaneous

Introduction to Acceleration

In non-uniform motion, the velocity of an object changes with time. This change in velocity can be a change in its magnitude (speed), a change in its direction, or both. Acceleration is the physical quantity that measures the rate of change of velocity.

Average Acceleration

The average acceleration ($\bar{a}$) over a time interval is defined as the change in velocity divided by the time interval. $$\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$$

• It is a vector quantity, and its direction is the same as the direction of the change in velocity ($\Delta v$).
• Its SI unit is meter per second squared ($m/s^2$ or $ms^{-2}$).
• The dimensional formula is $[M^0 L^1 T^{-2}]$.

Instantaneous Acceleration

The acceleration of an object at a particular instant of time is called instantaneous acceleration ($a$). It is defined as the limit of the average acceleration as the time interval $\Delta t$ becomes infinitesimally small.

Mathematically, it is the first derivative of the velocity with respect to time, and the second derivative of position with respect to time: $$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2x}{dt^2}$$

Graphical Interpretation

• On a velocity-time (v-t) graph, the instantaneous acceleration at any point is the slope of the tangent to the curve at that point.
• A positive slope indicates positive acceleration.
• A negative slope indicates negative acceleration (retardation).
• A zero slope (horizontal line) indicates zero acceleration (constant velocity).

Positive and Negative Acceleration

• If acceleration is in the same direction as velocity (both positive or both negative), the object speeds up.
• If acceleration is in the opposite direction to velocity (one positive, one negative), the object slows down. This is often called deceleration or retardation.

Example 1:

The position of a particle moving along the x-axis is given by $x(t) = t^3 - 6t^2 + 9t$, where x is in meters and t is in seconds. Find: (a) The velocity at t = 2 s. (b) The acceleration at t = 4 s. (c) The time intervals when the particle is speeding up and slowing down.

Solution: First, we find the expressions for velocity and acceleration by differentiation. Velocity: $v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9$. Acceleration: $a(t) = \frac{dv}{dt} = 6t - 12$.

(a) Velocity at t = 2 s: $v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3\ m/s$.

(b) Acceleration at t = 4 s: $a(4) = 6(4) - 12 = 24 - 12 = 12\ m/s^2$.

(c) Speeding up vs. Slowing down: We need to find the signs of v(t) and a(t).

• $a(t) = 6t - 12$ is zero when $t=2s$. It is negative for $t<2s$ and positive for $t>2s$.
• $v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$ is zero when $t=1s$ and $t=3s$. It is positive for $t<1$ and $t>3$, and negative for $1 < t < 3$.

Let's analyze the intervals:

• 0 < t < 1s: v > 0, a < 0. Signs are opposite $\implies$ Slowing down.
• 1 < t < 2s: v < 0, a < 0. Signs are the same $\implies$ Speeding up.
• 2 < t < 3s: v < 0, a > 0. Signs are opposite $\implies$ Slowing down.
• t > 3s: v > 0, a > 0. Signs are the same $\implies$ Speeding up.