2.2 Acceleration and deceleration  (Page 3/4)

Clearly, the phrases like “increase or decrease in velocity” or “increase or decrease in acceleration” are correct only when motion is "unidirectional" and in "positive" reference direction. Only in this restricted context, we can say that acceleration and velocity are increasing or decreasing. In order to be consistent with algebraic meaning, however, we may prefer to associate relative measure (increase or decrease) with magnitude of the quantity and not with the vector quantity itself.

Deceleration

Acceleration is defined strictly as the time rate of change of velocity vector. Deceleration, on the other hand, is acceleration that causes reduction in "speed". Deceleration is not opposite of acceleration. It is certainly not negative time rate of change of velocity. It is a very restricted term as explained below.

We have seen that speed of a particle in motion decreases when component of acceleration is opposite to the direction of velocity. In this situation, we can say that particle is being decelerated. Even in this situation, we can not say that deceleration is opposite to acceleration. Here, only a component of acceleration is opposite to velocity – not the entire acceleration. However, if acceleration itself (not a component of it) is opposite to velocity, then deceleration is indeed opposite to acceleration.

If we consider motion in one dimension, then the deceleration occurs when signs of velocity and acceleration are opposite. A negative velocity and a positive acceleration mean deceleration; a positive velocity and a negative acceleration mean deceleration; a positive velocity and a positive acceleration mean acceleration; a negative velocity and a negative acceleration mean acceleration.

Take the case of projectile motion of a ball. We study this motion as two equivalent linear motions; one along x-direction and another along y-direction.

Parabolic motion

For the upward flight, velocity is positive and acceleration is negative. As such the projectile is decelerated and the speed of the ball in + y direction decreases (deceleration). For downward flight from the maximum height, velocity and acceleration both are negative. As such the projectile is accelerated and the speed of the ball in - y direction increases (acceleration).

In the nutshell, we summarize the discussion as :

• Deceleration results in decrease in speed i.e magnitude of velocity.
• In one dimensional motion, the “deceleration” is defined as the acceleration which is opposite to the velocity.

Acceleration and deceleration

Problem : The velocity of a particle along a straight line is plotted with respect to time as shown in the figure. Find acceleration of the particle between OA and CD. What is acceleration at t = 0.5 second and 1.5 second. What is the nature of accelerations in different segments of motion? Also investigate acceleration at A.

Velocity – time plot

Solution : Average acceleration between O and A is given by the slope of straight line OA :

$$\begin{array}{l}{a}_{\mathrm{OA}}=\frac{v_2-v_1}{t}=\frac{0.1-0}{1}=0.1m/s^2\end{array}$$

Average acceleration between C and D is given by the slope of straight line CD :

$$\begin{array}{l}{a}_{\mathrm{CD}}=\frac{v_2-v_1}{t}=\frac{0-(-0.1)}{1}=0.1m/s^2\end{array}$$

Accelerations at t = 0.5 second and 1.5 second are obtained by determining slopes of the curve at these time instants. In the example, the slopes at these times are equal to the slope of the lines OA and AB.

Instantaneous acceleration at t = 0.5 s :

$$\begin{array}{l}{a}_{0.5}=a_{\mathrm{OA}}=0.1m/s^2\end{array}$$

Instantaneous acceleration at t = 1.5 s :

$$\begin{array}{l}{a}_{1.5}=a_{\mathrm{AB}}=\frac{v_2-v_1}{t}=\frac{0-0.1}{1}=-0.1m/s^2\end{array}$$

We check the direction of velocity and acceleration in different segments of the motion in order to determine deceleration. To enable comparision, we determine directions with respect to the assumed positive direction of velocity. In OA segment, both acceleration and velocity are positive (hence particle is accelerated). In AB segment, acceleration is negative, but velocity is positive (hence particle is decelerated). In BC segment, both acceleration and velocity are negative (hence particle is accelerated). In CD segment, acceleration is positive but velocity is negative (hence particle is decelerated).

Alternatively, the speed increases in segment OA and BC (hence acceleration); decreases in segments AB and CD (hence deceleration).

We note that it is not possible to draw an unique tangent at point A. We may draw infinite numbers of tangent at this point. In other words, limit of average acceleration can not be evaluated at A. Acceleration at A, therefore, is indeterminate.

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