-
EQUATION FOR VELOCITY- TIME RELATION
INTRODUCTION TO EQUATIONS OF MOTION
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion, and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:
v = u + at ______ (8.5)
s = ut + ½ at2 ______ (8.6)
2 as = v2 – u2 ______ (8.7)
where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation. Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by the graphical method.
EQUATION FOR VELOCITY-TIME RELATION
Figure 8.8: Velocity-time graph to obtain the equations of motion
Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Fig. 8.8 (similar to Fig. 8.6, but now with u \(\ne\) 0). From this graph, you can see that the initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t.
Let us draw AD parallel to OC. From the graph, we observe that
BC = BD + DC = BD + OA
Substituting BC = v and OA = u,
we get v = BD + u
or BD = v – u ________ (8.8)
From the velocity-time graph (Fig. 8.8),the acceleration of the object is given by
\(a=\frac{change \,\,in\,\,velocity}{time\,\,taken}\\ =\frac{BD}{AD}=\frac{BD}{OC}\)
Substituting OC = t, we get
\(A=\frac{BD}{t}\)
or BD = at ________ (8.9)
Using Eqs. (8.8) and (8.9) we get
v = u + at
Source: This topic is taken from NCERT TEXTBOOK
-
EQUATION FOR POSITION- TIME RELATION
EQUATION FOR POSITION-TIME RELATION
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Figure 8.8: Velocity-time graph to obtain the equations of motion
Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABD
\(=OA\times OC+\frac{1}{2}(AD\times BD)\) ______(8.10)
Substituting OA = u, OC = AD = t and BD = at,
we get
\(s=u\times t+\frac{1}{2}(t\times at) \\ or\,\, s=ut+\frac{1}{2}at^2\)
Illustration 8.7:
The brakes applied to a car produce an acceleration of 6 ms-2 in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time.
Sol:
We have been given
a = – 6 ms–2 ; t = 2 s and v = 0 ms–1. From Eq. (8.5) we know that
v = u + at
0 = u + (– 6 ms–2) × 2 s
or u = 12 ms–1 . From Eq. (8.6) we get
\(s=ut+\frac{1}{2}at^2\)
\(=(12ms^{-1})\times(2s)+\frac{1}{2}(-6ms^{-2})(2s)^2\)
= 24 m – 12 m
= 12 m
Thus, the car will move 12 m before it stops after the application of brakes.
Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road?
Source: This topic is taken from NCERT TEXTBOOK
-
EQUATION FOR POSITION-VELOCITY RELATION
EQUATION FOR POSITION-VELOCITY RELATION
From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,
Figure 8.8: Velocity-time graph to obtain the equations of motion
s = area of the trapezium OABC
= \(= \frac{(OA+BC)\times OC}{2}\)
Substituting OA = u, OC = AD = t and BD = at,
we get
\(s=\frac{(u+v)t}{2}\) ______(8.11)
From the velocity-time relation (Eq. 8.6), we get
\(t=\frac{(v-u)}{a}\) ______(8.12)
Using Eqs. (8.11) and (8.12) we have
\(= \frac{(v+u)\times (v-u)}{2a}\)
or 2as = v2 – u2
Illustration 8.5:
A train starting from rest attains a velocity of 72 kmh–1 in 5 minutes. Assuming that the acceleration is uniform, find
(i) the acceleration and
(ii) the distance travelled by the train for attaining this velocity.
Sol:
We have been given
u = 0 ; v = 72 kmh–1 = 20 ms-1 and
t = 5 minutes = 300 s.
From Eq. (8.5) we know that
\(a=\frac{v-u}{t}\\ =\frac{(20\,ms^{-1})-(0\,ms^{-1})}{300s}\\ =\frac{1}{15}ms^{-2}\)
From Eq. (8.7) we have 2 a s = v2 – u2 = v2 – 0
Thus,
\(s=\frac{v^2}{2a}\\ =\frac{(20ms^{-1})^2}{2\times (1/15)ms^{-2}}\\ =300m =3km\)
The acceleration of the train is \(\frac{1}{15}ms^{-2}\) and the distance travelled is 3 km.
Illustration 8.6:
A car accelerates uniformly from 18 kmh-1 to 36 kmh-1 in 5 s. Calculate
(i) the acceleration and
(ii) the distance covered by the car in that time
Sol:
We are given that
u = 18 kmh–1 = 5 ms–1
v = 36 kmh–1 = 10 ms–1 and
t = 5 s .
(i) From Eq. (8.5) we have
\(a=\frac{v-u}{t}\\ =\frac{(10ms^{-1}-5ms^{-1})}{5s}\\ =1\,ms^{-2} \)
(ii) From Eq. (8.6) we have
\(s=ut+\frac{1}{2}at^2\\ =5ms^{-1}\times5s+\frac{1}{2}\times 1ms^{-2}\times (5s)^2\\ =25m+12.5m =37.5m\)
The acceleration of the car is 1 ms and the distance covered is 37.5 m.
Questions
1. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find ,
(i) the speed acquired
(ii) the distance travelled.
2. A train is travelling at a speed of 90 kmh–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 ms-2. Find how far the train will go before it is brought to rest.
3. A trolley, while going down an inclined plane, has an acceleration of 2cms-2. What will be its velocity 3 s after the start?
4. A racing car has a uniform acceleration of 4 ms-2. What distance will it cover in 10 s after start?
5. A stone is thrown in a vertically upward direction with a velocity of 5 ms-1. If the acceleration of the stone during its motion is 10 ms–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Source: This topic is taken from NCERT TEXTBOOK