MOTION IN A STRAIGHT LINE

and x ( t ) = . t . The sixth column lists the difference ∆ x = x ( t ) – x ( t ) and the last column gives the ratio of ∆ x and ∆ t , i.e. the average velocity corresponding to the value of ∆ t listed in the first column.

We see from Table . that as we decrease the value of ∆ t from . s to . s, the value of the average velocity approaches the limiting value .

m s – which is the value of velocity at t = . s, i.e. the value of dx dt at t = . s.

In this manner, we can calculate velocity at each instant for motion of the car. The graphical method for the determination of the instantaneous velocity is always not a convenient method. For this, we must carefully plot the position–time graph and calculate the value of average velocity as ∆ t becomes smaller and smaller. It is easier to calculate the value of velocity at different instants if we have data of positions at different instants or exact expression for the position as a function of time.

Then, we calculate ∆ x / ∆ t from the data for decreasing the value of ∆ t and find the limiting value as we have done in Table . or use differential calculus for the given expression and calculate dx dt at different instants as done in the following example. Example . The position of an object moving along x-axis is given by x = a + bt where a = .

m , b = . m s – and t is measured in seconds. What is its velocity at t = s and t = . s .

What is the average velocity between t = . s and t = . s ? Answer In notation of differential calculus, the velocity is - dx bt 2b t = .

t m s dt dt At t = s, v = m s – and at t = . s, v = m s - . . .

. . Average velocity – . .

b b b × - . . = m s × Note that for uniform motion, velocity is the same as the average velocity at all instants . Instantaneous speed or simply speed is the magnitude of velocity.

For example, a velocity of + . m s – and a velocity of – . m s – — both have an associated speed of . m s - .

It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so ? . ACCELERATION The velocity of an object, in general, changes during its course of motion.

How to describe this change? Should it be described as the rate of change in velocity with distance or with time ? This was a problem even in Galileo’s time. It was first thought that this change could be described by the rate of change of velocity with distance.

But, through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. On the other hand, the change in velocity with distance is not constant – it decreases with the increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time. The average acceleration a over a time interval is defined as the change of velocity divided by the time interval : = ∆ ( .

) where v and v are the instantaneous velocities or simply velocities at time t and t . It is the average change of velocity per unit time. The SI unit of acceleration is m s – . On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to ( v , t ) and ( v , t ).

Fig. . Velocity–time graph for motions with constant acceleration. (a) Motion in positive direction with positive acceleration, (b) Motion in positive direction with negative acceleration, (c) Motion in negative direction with negative acceleration, (d) Motion of an object with negative acceleration that changes direction at time t .

Between times to t , it moves in positive x - direction and between t and t it moves in the opposite direction . Instantaneous acceleration is defined in the same way as the instantaneous velocity : lim ( . ) The acceleration at an instant is the slope of the tangent to the v–t curve at that instant. Since velocity is a quantity having both magnitude and direction, a change in velocity may involve either or both of these factors.

Acceleration, therefore, may result from a change in speed (magnitude), a change in direction or changes in both. Like velocity, acceleration can also be positive, negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. .

(a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval.

If the velocity of an object is v o at t = and v at time t , we have or , v a t ( . ) Fig. . Position-time graph for motion with (a) positive acceleration; (b) negative acceleration, and (c) zero acceleration .

Let us see how velocity-time graph looks like for some simple cases. Fig. . shows velocity- time graph for motion with constant acceleration for the following cases : (a) An object is moving in a positive direction with a positive acceleration.

(b) An object is moving in positive direction with a negative acceleration. (c) An object is moving in negative direction with a negative acceleration. (d) An object is moving in positive direction till time t , and then turns back with the same negative acceleration. An interesting feature of a velocity-time graph for any moving object is that the area under the curve represents the displacement over a given time interval .

A general proof of this statement requires use of calculus. We can, however, see that it is true for the simple case of an object moving with constant velocity u . Its velocity-time graph is as shown in Fig. .

. Fig. . Area under v–t curve equals displacement of the object over a given time interval .

The v-t curve is a straight line parallel to the time axis and the area under it between t = and t = T is the area of the rectangle of height u and base T . Therefore, area = u × T = uT which is the displacement in this time interval. How come in this case an area is equal to a distance? Think!

Note the dimensions of quantities on the two coordinate axes, and you will arrive at the answer. Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp kinks at some points implying that the functions are not differentiable at these points. In any realistic situation, the functions will be differentiable at all points and the graphs will be smooth . What this means physically is that acceleration and velocity cannot change values abruptly at an instant.

Changes are always continuous. . KINEMATIC E Q UATIONS FOR UNIFORMLY ACCELERATED MOTION For uniformly accelerated motion, we can derive some simple equations that relate displacement ( x ), time taken ( t ), initial velocity ( v ), final velocity ( v ) and acceleration ( a ). Equation ( .

) already obtained gives a relation between final and initial velocities v and v of an object moving with uniform acceleration a : v = v + at ( . ) This relation is graphically represented in Fig. . .

The area under this curve is : Area between instants and t = Area of triangle ABC + Area of rectangle OACD v v t + v t Fig. . Area under v-t curve for an object with uniform acceleration . As explained in the previous section, the area under v-t curve represents the displacement.

Therefore, the displacement x of the object is : v v t + v t ( . ) But a t Therefore, a t + v t or, ( . ) Equation ( . ) can also be written as v + v t v t ( .7a) where, (constant acceleration only) ( .7b) Equations ( .7a) and ( .7b) mean that the object has undergone displacement x with an average velocity equal to the arithmetic average of the initial and final velocities.

From Eq. ( . ), t = ( v – v ) /a . Substituting this in Eq.

( .7a), we get            = ax ( . ) This equation can also be obtained by substituting the value of t from Eq. ( . ) into Eq.

( . ). Thus, we have obtained three important equations : ax ( .9a) connecting five quantities v , v, a, t and x . These are kinematic equations of rectilinear motion for constant acceleration.

The set of Eq. ( .9a) were obtained by assuming that at t = , the position of the particle, x is . We can obtain a more general equation if we take the position coordinate at t = as non- zero, say x . Then Eqs.

( .9a) are modified (replacing x by x – x ) to : ( .9b) ( a x ( .9c) Example . Obtain equations of motion for constant acceleration using method of calculus. Answer By definition d v = a d t Integrating both sides