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Instantaneous Acceleration

When an object is moving, then it changes its speed and thus there is a change in velocity. An acceleration is the rate of velocity change in the given time period. It is a vector quantity with magnitude and direction and its changes with changing the direction of the velocity. It is positive with increasing velocity while negative with decreasing velocity. The negative directed acceleration is called retardation and deceleration.

When velocity is constant for a particular period of time then the object is in constant acceleration position. If we talk about one dimension, the rate at which objects get slower speed or speed up is equal to the acceleration. In case of instantaneous acceleration, it is the change at a particular moment. The object with different acceleration at different moments of time then the object is in a variable acceleration state. This is instantaneous acceleration. Here we are discussing more about on instantaneous acceleration with its mathematical formula and problem based on it.

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Acceleration

We know that Velocity is the rate of change of distance with respect to time. So what can we call if there is rate of change of Velocity? Its nothing but Acceleration.

The Acceleration is the rate of change of velocity with respect to time.

The Acceleration can also be defined mathematically by,
a = $\frac{v}{t}$,Where,
a = acceleration of the moving body
v = velocity of the moving body
t = time

Instantaneous Acceleration Formula

To know what is Instantaneous Acceleration, let us consider acceleration of a body at a particular instant of time. So, at that instant of time, let us find its acceleration, that will be its Instantaneous Acceleration. The acceleration of the moving body at any instant of time is defined as its instantaneous acceleration.

The Instantaneous acceleration equation is given by,
ains = $\lim_{t \to 0}$ $\frac {dv}{dt}$
= $\lim_{t \to 0}$ $\frac{\mathrm{d^{2}x}}{\mathrm{d} t^{2}}$.

The acceleration of the body can be defined in terms of Average Acceleration. The Average Acceleration is defined as the difference between velocities at two points divided by the time taken to travel the distance between them.

The Average acceleration is given by,
aavg = $\frac{( v_{2} – v_{1})} {( t_{2} – t_{1} )}$ In order to find the instantaneous acceleration, we find the average acceleration for both initial and final point, when we reach that interval of time.

How to Find Instantaneous Acceleration?

If the body is moving with the Constant velocity then its instantaneous acceleration is zero, since acceleration is measure of how fast a body is changing its velocity. For the body moving with varying velocity, the instantaneous acceleration can be obtained by using the above formula of the instantaneous acceleration.
Given below are some numerical problems which will helps us have a better understanding of it.

Solved Examples

Question 1: A diesel locomotive is moving on the rail track. The distance traveled by it is defined by the following equation: x = 4t2 + 4t + 12,
a. Find the instantaneous velocity at time t = 10 seconds?
b. Find the instantaneous acceleration at time t = 18 seconds?
Solution:

a. The instantaneous velocity is defined by,
V$_{ins}$ = $\lim_{t \to10}$ $\frac{dx}{dt}$,
V$_{ins}$ = $\lim_{t \to10}$ $\frac{d(4t^{2} + 4t + 12)}{dt}$,
V$_{ins}$ = $\lim_{t \to 10}$ (8t + 4),
V$_{ins}$ = 8 $\times$ 10 + 4
V$_{ins}$ = 84 m/s.

Similarly,
b. The instantaneous acceleration is defined by,
a$_{ins}$ = $\lim_{t \to 0}$ $\frac{d^{2} x}{dt^{2}}$,
= $\lim_{t \to 0}$ $\frac{d^{2}(4t^{2} + 4t + 12)}{dt^{2}}$,
= $\lim_{t \to 0}$ {8t + 0}
= $\lim_{t \to 0}$ (8 $\times$ 1),
= 8 m/s$^{2}$.

Question 2: Find the instantaneous acceleration of a truck, at time t = 6 seconds, moving with the velocity v given by, v = 3t3 + 4t2 + 5t + 6.
Solution:

Let us first find the acceleration by finding the first derivative of the velocity equation,
a = $\frac{d (3t^{3} + 4t^{2} + 5t + 6)}{dt}$ ---------(1)
a = 9t$^{2}$ + 8t + 5 ---------(2)
So the instantaneous acceleration at t = 6 seconds is given by;
a = $\lim_{t \to 6}$ (9t$^{2}$ + 8t + 5),
a = 9 $\times$ 36+ 8 $\times$ 6 + 5
= 324 + 48 + 5
= 377 m/s$^{2}$.

Question 3: A car is moving on a linear track with the varying acceleration. Its velocity is given by, v = 2t2 + 2t + 4. Find the instantaneous acceleration of the car at t = 8 seconds?
Solution:

The instantaneous acceleration is given by, a$_{ins}$ = $\lim_{t \to 0}$ $\frac{dv}{dt}$,
a$_{ins}$ = $\lim_{t \to 8}$ $\frac{d(2t^{2} + 2t + 4)}{dt}$,
a$_{ins}$ = $\lim_{t \to 8}$ (4t + 2),
a$_{ins}$ = 4 $\times$ 8 + 2,
a$_{ins}$ = 34 m/s$^{2}$.
The acceleration in this is varying, since the acceleration is a function of time (t).

Instantaneous Acceleration Graph 