-
ACCELERATION
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 \(\bar a\) over a time interval is defined as the change of velocity divided by the time interval :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaGaamODamaaBaaaleaacaaIYaaabeaa % kiabgkHiTiaadAhadaWgaaWcbaGaaGymaaqabaaakeaacaWG0bWaaS % baaSqaaiaaikdaaeqaaOGaeyOeI0IaamiDamaaBaaaleaacaaIXaaa % beaaaaGccqGH9aqpdaWcaaqaaiabgs5aejaadAhaaeaacqGHuoarca % WG0baaaaaa!4763! \overline a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}} = \frac{{\Delta v}}{{\Delta t}}\)
where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1.It is the average change of velocity per unit time.The SI
unit of acceleration is m s–2 .
On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to (\(v\) 2 , t2) and (\(v\) 1 , t 1). The average acceleration for velocity-time graph shown in Fig. 3.7 for different time intervals 0 s - 10 s, 10 s – 18 s, and 18 s – 20 s are :
0 s -10 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIYaGaaGinaiabgkHi % TiaaicdaaiaawIcacaGLPaaacaWGTbGaam4CamaaCaaaleqabaGaey % OeI0IaaGymaaaaaOqaamaabmaabaGaaGymaiaaicdacqGHsislcaaI % WaaacaGLOaGaayzkaaGaam4CaaaacqGH9aqpcaaIYaGaaiOlaiaais % dacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaa!4D05! \overline a = \frac{{\left( {24 - 0} \right)m{s^{ - 1}}}}{{\left( {10 - 0} \right)s}} = 2.4m{s^{ - 2}}\)
10 s-18 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIYaGaaGinaiabgkHi % TiaaikdacaaI0aaacaGLOaGaayzkaaGaamyBaiaadohadaahaaWcbe % qaaiabgkHiTiaaigdaaaaakeaadaqadaqaaiaaigdacaaI4aGaeyOe % I0IaaGymaiaaicdaaiaawIcacaGLPaaacaWGZbaaaiabg2da9iaaic % dacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaa!4D16! \overline a = \frac{{\left( {24 - 24} \right)m{s^{ - 1}}}}{{\left( {18 - 10} \right)s}} = 0m{s^{ - 2}}\)
18 s-20 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIWaGaeyOeI0IaaGOm % aiaaisdaaiaawIcacaGLPaaacaWGTbGaam4CamaaCaaaleqabaGaey % OeI0IaaGymaaaaaOqaamaabmaabaGaaGOmaiaaicdacqGHsislcaaI % XaGaaGioaaGaayjkaiaawMcaaiaadohaaaGaeyypa0JaeyOeI0IaaG % ymaiaaikdacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaa % aaa!4E01! \overline a = \frac{{\left( {0 - 24} \right)m{s^{ - 1}}}}{{\left( {20 - 18} \right)s}} = - 12m{s^{ - 2}}\)
fig 3.8. Acceleration as a function of time for motion represented in Fig. 3.3.
Instantaneous acceleration is defined in the same way as the instantaneous velocity :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9maaxacabaGaeyiLdqKaamiDaiabgkziUkaaicdaaSqabeaaciGG % SbGaaiyAaiaac2gaaaGcdaWcaaqaaiabgs5aejaadAhaaeaacqGHuo % arcaWG0baaaiabg2da9maalaaabaGaamizaiaadAhaaeaacaWGKbGa % amiDaaaaaaa!49BA! a = \mathop {\Delta t \to 0}\limits^{\lim } \frac{{\Delta v}}{{\Delta t}} = \frac{{dv}}{{dt}}\)
The acceleration at an instant is the slope of the tangent to the v–t curve at that instant. For the v–t curve shown in Fig. 3.7, we can obtain acceleration at every instant of time.The resulting a – t curve is shown in Fig. 3.8. We see that the acceleration is nonuniform over the period 0 s to 10 s. It is zero between 10 s and18 s and is constant with value –12 m s–2 between 18 s and 20 s.When the acceleration is uniform, obviously, it equals the average acceleration over that period.
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. 3.9 (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. As an exercise, identify in Fig. 3.3, the regions of the curve that correspond to these three cases.
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 \(v\)0 at t = 0 and \(v\) at time t,
we have,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaGaamODaiabgkHiTiaadAhadaWgaaWc % baGaaGimaaqabaaakeaacaWG0bGaeyOeI0IaaGimaaaacaWGVbGaam % OCaiaacYcacaWG2bGaeyypa0JaamODamaaBaaaleaacaaIWaaabeaa % kiabgUcaRiaadggacaWG0baaaa!47BE! \overline a = \frac{{v - {v_0}}}{{t - 0}}or,v = {v_0} + at\)
fig 3.9. 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. 3.10 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, for example the motion of the car in Fig. 3.3 betweent = 0 s and t = 10 s.
(b)An object is moving in positive direction with a negative acceleration, for example, motion of the car in Fig 3.3 between t = 18 s and 20 s.
(c)An object is moving in negative direction with a negative acceleration, for example the motion of a car moving from O in Fig. 3.1 in negative x-direction with increasing speed.
(d)An object is moving in positive direction till time t1, and then turns back with the same negative acceleration, for example the motion of a car from point O to point Q in Fig. 3.1 till time t1 with decreasing speed and turning back and moving 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
fig 3.10.
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 timet.
Between times 0 to t1, its moves in positive x - direction and between t1 t2 it
moves in the opposite direction.
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. 3.11.
fig 3.11. 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 = 0 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.
-
ACCELERATION
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 \(\bar a\) over a time interval is defined as the change of velocity divided by the time interval :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaGaamODamaaBaaaleaacaaIYaaabeaa % kiabgkHiTiaadAhadaWgaaWcbaGaaGymaaqabaaakeaacaWG0bWaaS % baaSqaaiaaikdaaeqaaOGaeyOeI0IaamiDamaaBaaaleaacaaIXaaa % beaaaaGccqGH9aqpdaWcaaqaaiabgs5aejaadAhaaeaacqGHuoarca % WG0baaaaaa!4763! \overline a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}} = \frac{{\Delta v}}{{\Delta t}}\)
where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1.It is the average change of velocity per unit time.The SI
unit of acceleration is m s–2 .
On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to (\(v\) 2 , t2) and (\(v\) 1 , t 1). The average acceleration for velocity-time graph shown in Fig. 3.7 for different time intervals 0 s - 10 s, 10 s – 18 s, and 18 s – 20 s are :
0 s -10 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIYaGaaGinaiabgkHi % TiaaicdaaiaawIcacaGLPaaacaWGTbGaam4CamaaCaaaleqabaGaey % OeI0IaaGymaaaaaOqaamaabmaabaGaaGymaiaaicdacqGHsislcaaI % WaaacaGLOaGaayzkaaGaam4CaaaacqGH9aqpcaaIYaGaaiOlaiaais % dacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaa!4D05! \overline a = \frac{{\left( {24 - 0} \right)m{s^{ - 1}}}}{{\left( {10 - 0} \right)s}} = 2.4m{s^{ - 2}}\)
10 s-18 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIYaGaaGinaiabgkHi % TiaaikdacaaI0aaacaGLOaGaayzkaaGaamyBaiaadohadaahaaWcbe % qaaiabgkHiTiaaigdaaaaakeaadaqadaqaaiaaigdacaaI4aGaeyOe % I0IaaGymaiaaicdaaiaawIcacaGLPaaacaWGZbaaaiabg2da9iaaic % dacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaa!4D16! \overline a = \frac{{\left( {24 - 24} \right)m{s^{ - 1}}}}{{\left( {18 - 10} \right)s}} = 0m{s^{ - 2}}\)
18 s-20 s=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaWaaeWaaeaacaaIWaGaeyOeI0IaaGOm % aiaaisdaaiaawIcacaGLPaaacaWGTbGaam4CamaaCaaaleqabaGaey % OeI0IaaGymaaaaaOqaamaabmaabaGaaGOmaiaaicdacqGHsislcaaI % XaGaaGioaaGaayjkaiaawMcaaiaadohaaaGaeyypa0JaeyOeI0IaaG % ymaiaaikdacaWGTbGaam4CamaaCaaaleqabaGaeyOeI0IaaGOmaaaa % aaa!4E01! \overline a = \frac{{\left( {0 - 24} \right)m{s^{ - 1}}}}{{\left( {20 - 18} \right)s}} = - 12m{s^{ - 2}}\)
fig 3.8. Acceleration as a function of time for motion represented in Fig. 3.3.
Instantaneous acceleration is defined in the same way as the instantaneous velocity :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9maaxacabaGaeyiLdqKaamiDaiabgkziUkaaicdaaSqabeaaciGG % SbGaaiyAaiaac2gaaaGcdaWcaaqaaiabgs5aejaadAhaaeaacqGHuo % arcaWG0baaaiabg2da9maalaaabaGaamizaiaadAhaaeaacaWGKbGa % amiDaaaaaaa!49BA! a = \mathop {\Delta t \to 0}\limits^{\lim } \frac{{\Delta v}}{{\Delta t}} = \frac{{dv}}{{dt}}\)
The acceleration at an instant is the slope of the tangent to the v–t curve at that instant. For the v–t curve shown in Fig. 3.7, we can obtain acceleration at every instant of time.The resulting a – t curve is shown in Fig. 3.8. We see that the acceleration is nonuniform over the period 0 s to 10 s. It is zero between 10 s and18 s and is constant with value –12 m s–2 between 18 s and 20 s.When the acceleration is uniform, obviously, it equals the average acceleration over that period.
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. 3.9 (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. As an exercise, identify in Fig. 3.3, the regions of the curve that correspond to these three cases.
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 \(v\)0 at t = 0 and \(v\) at time t,
we have,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGHbaaaiabg2da9maalaaabaGaamODaiabgkHiTiaadAhadaWgaaWc % baGaaGimaaqabaaakeaacaWG0bGaeyOeI0IaaGimaaaacaWGVbGaam % OCaiaacYcacaWG2bGaeyypa0JaamODamaaBaaaleaacaaIWaaabeaa % kiabgUcaRiaadggacaWG0baaaa!47BE! \overline a = \frac{{v - {v_0}}}{{t - 0}}or,v = {v_0} + at\)
fig 3.9. 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. 3.10 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, for example the motion of the car in Fig. 3.3 betweent = 0 s and t = 10 s.
(b)An object is moving in positive direction with a negative acceleration, for example, motion of the car in Fig 3.3 between t = 18 s and 20 s.
(c)An object is moving in negative direction with a negative acceleration, for example the motion of a car moving from O in Fig. 3.1 in negative x-direction with increasing speed.
(d)An object is moving in positive direction till time t1, and then turns back with the same negative acceleration, for example the motion of a car from point O to point Q in Fig. 3.1 till time t1 with decreasing speed and turning back and moving 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
fig 3.10.
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 timet.
Between times 0 to t1, its moves in positive x - direction and between t1 t2 it
moves in the opposite direction.
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. 3.11.
fig 3.11. 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 = 0 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.