(5)
Q.5 Answer the following questions :
Derive the equations of motion by graphical method for an object having uniformly
accelerated motion.
in writing.
Answers
Explanation:
Equations of motion by graphical method:
Consider an object moving along a straight line with initial velocity 'u' and uniform acceleration a.
Suppose, it travels distance s in time t.
As shown in Fig. 8.19, its velocity-time graph is straight line.
Here,
OA = ED = u
OC = EB = v and
OE = t = AD.
1. Equation for velocity-time relation.
Acceleration is given by slope of velocity-time graph AB
space space straight a space equals space DB over AD space equals space DB over OE space equals space fraction numerator EB minus ED over denominator OE end fraction space equals space fraction numerator straight v minus straight u over denominator straight t end fraction
i. e. comma space space space space straight v minus straight u space equals space at space
rightwards double arrow space space space space space space box enclose straight v space equals space straight u space plus space at end enclose
Hence, the first equation of motion is proved.
2. Equation for position-time relation:
From the first part, we have
Acceleration comma space straight a space equals space DB over AD space equals DB over straight t
rightwards double arrow space space space space space space space space space space space space space space space space space DB space equals space at
Distance space travelled space by space the space object space in space time space straight t space is space g i v e n space b y comma space
space straight s space equals space Area space of space the space trapezium space OABE
space space space space equals space Area space of space rectangle space OADE space plus space Area space of space triangle space ADB
space space space space space space space space space space equals space OA space cross times space OE space plus space 1 half DB cross times AD
space space space space space space space space space equals space ut space plus space 1 half at cross times straight t
rightwards double arrow space straight s space equals space u t space plus space 1 half a t squared
This proves the second equation of motion.
3. Equation for position-velocity relation:
The distance travelled by object in time t is,
s = Area of trapezium OABE
equals 1 half open parentheses E B plus O A close parentheses cross times O E
equals space 1 half open parentheses E B plus E D close parentheses cross times O E space
Acceleration, a = slope of velocity-time graph AB.
space a equals DB over AD equals fraction numerator E B minus E D over denominator O E end fraction
rightwards double arrow space O E space equals space fraction numerator E B space minus space E D over denominator a end fraction
therefore space D i s tan c e space t r a v e l l e d comma space s space equals space 1 half open parentheses E B plus E D close parentheses cross times fraction numerator left parenthesis E B minus E D right parenthesis over denominator a end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 1 over denominator 2 a end fraction open parentheses E B squared space minus space E D squared close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1 over denominator 2 a end fraction open parentheses v squared minus u squared close parentheses space
i. e. comma space space space space space space space space space space space v squared space minus space u squared space equals space 2 a s
Hence, the third equation of motion is also proved.
Answer:
Motion with constant acceleration or uniformly accelerated motion is that in which velocity changes at the same rate throughout motion.
When the acceleration of the moving object is constant its average acceleration and instantaneous acceleration are equal. Thus from eq. 5 we have
Kinematic equations for uniformly accelerated Motion
Let v0 be the velocity at initial time t=0 and v be the velocity of object at some other instant of time say at t2=t then above eq. 7 becomes
a
=
v
−
v
0
t
−
0
or , v=v0+at (8)
Graphically this relation is represented in figure 8 given below.
graph between position and time in rectilinear motion with constant acceleration
Thus from the graph it can be seen clearly that velocity v at time t is equal to the velocity v0 at time t=0 plus the change in velocity (at).
In the same way average velocity can be written as
v
a
v
g
=
x
−
x
0
t
−
0
where x0 is the position of object at time t=0 and vavg is the average velocity between time t=0 to time t.The above equation then gives
x=x0+vavgt (9)
Now we know that area under v-t curve represent the displacement
So,
x
−
x
0
=
1
2
(
v
−
v
0
)
t
+
v
0
t
as ( Area of Triangle + Area of Rectangle)
x
−
x
0
=
v
+
v
0