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MOTION IN A STRAIGHT LINE 15 corresponding values of x, i.e. x (t ) = 0.08 t3 =a +16b – a – 4b = 6.0 ×b 1 1 2.0 and x (t ) = 0.08 3. The sixth column lists the -1 2 t2 = 6.0×2.5 =15 m s ⊳ difference ∆x = x (t ) 2 x (t ) 1nd the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting –1 –1 –1 + 24.0 m s and a velocity of – 24.0 m s -1— value 3.84 m s which is the value of velocity at both have an associated speed of 24.0 m s . It dx should be noted that though average speed over t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt manner, we can calculate velocity at each magnitude of the average velocity, instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a convenient method. For this, we must carefully 2.3 ACCELERATION plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the increasing distance of fall. This led to the ⊳Example 2.1 The position of an object 2 concept of acceleration as the rate of change moving along x-axis is given by x = a + bt of velocity with time. where a = 8.5 m, b = 2.5 m s –2and t is Theaverageacceleration a overatimeinterval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? Answer In notation of differential calculus, the a v 2 v 1 ∆v (2.2) = t – t = ∆t velocity is 2 1 where v and v are the instantaneous velocities v = dx = d a + bt2 =2b t = 5.0 t m s-1 2 1 dt dt( ) or simply velocities at time t a2d t .1It is the –1 average change of velocity per unit time. The SI At t = 0 s, v = 0 m s and at t = 2.0 s, unit of acceleration is m s .2 v = 10 m s .1 On a plot of velocity versus time, the average x 4.0 − x 2.0 acceleration is the slope of the straight line Average velocity = ( ) ( ) connecting the points corresponding to (v , t2) 2 4.0 − 2.0 and (v , t ). 1 1 Reprint 2026-27