Kinematic equations for uniformly accelerated Motion
Kinematic equations for uniformly accelerated Motion
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
Let v_{0} be the velocity at initial time t=0 and v be the velocity of object at some other instant of time say at t_{2}=t then above eq. 7 becomes
$a=\frac{v - v_0}{t - 0}$
or ,
v=v_{0}+at (8)
Graphically this relation is represented in figure 8 given below.
Thus from the graph it can be seen clearly that velocity v at time t is equal to the velocity v_{0} at time t=0 plus the change in velocity (at).
In the same way average velocity can be written as
$v_{avg}=\frac{x - x_0}{t - 0}$
where x_{0} is the position of object at time t=0 and v_{avg} is the average velocity between time t=0 to time t.The above equation then gives
x=x_{0}+v_{avg}t (9)
Now we know that area under v-t curve represent the displacement
So,
$x -x_0=\frac{1}{2}(v -v_0)t + {v_0}t $ as ( Area of Triangle + Area of Rectangle)
$x -x_0=\frac{v+ v_0}{2}t $
From equation 9 and the above equation we cay say that
Now from eq. 8 we find v_{avg} = v_{0} + ½(at) (11)
putting this in eq. 9 we find
x = x_{0} + v_{0}t + ½(at^{2})
or,
x - x_{0} = v_{0}t + ½(at^{2}) (12)
this is the position time relation for object having uniformly accelerated motion.
From eq. 12 it is clear that an object at any time t has quadratic dependence on time, when it moves with constant acceleration along a straight line and x-t graph for such motion will be parabolic in natures shown below.
Equation 8 and 12 are basic equations for constant acceleration and these two equations can be combined to get yet another relation for x , v and a eliminating t so, from 8
putting this value of t in equation 12 and solving it we finally get,
v^{2} = (v_{0})^{2} + 2a ( x - x_{0} ) (13)
Thus from equation 13 we see that it is velocity displacement relation between velocities of object moving with constant acceleration at time t and t=0 and their corresponding positions at these intervals of time.
This relation 13 is helpful when we do not know time t.
Likewise we can also eliminate the acceleration between equation 8 and 12. Thus from equation 8
putting this value of a in equation 12 and solving it we finally get,
( x - x_{0} ) = ½ ( v_{0} + v ) t (14)
Same way we can also eliminate v_{0} using equation 8 and 12. Now from equation 8
v_{0} = v - at
putting this value of v_{0} in equation 12 and solving it we finally get,
( x - x_{0} ) = vt - ½ ( at^{2} ) (15)
thus equation 15 does not involve initial velocity v_{0}
Thus these basic equations 8 and 12 , and derived equations 13, 14, and 15 can be used to solve constant acceleration problems.
Obtaining Kinematics equations using Calculus Method
By definition,
$a= \frac{dv}{dt}$
dv=adt
Integrating Both sides
$\int dv =\int adt$
$\int dv =a\int dt$ ( as a is constant)
v-v_{0} = at
$v= v_0 + at$
Now,
$v= \frac{dx}{dt}$
dx= vdt
Integrating Both the sides
$\int dx =\int vdt$
or
$\int dx =\int (v_{0} + at)dt$
x - x_{0} = v_{0}t + ½(at^{2})
Now,
$a= \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx} $
or vdv=adx
Integrating Both sides
$\int vdv =\int adx$
$\frac{ v^{2} -v_{0}^{2}}{2} =a( x -x_{0})$
v^{2} = (v_{0})^{2} + 2a ( x - x_{0} )
So we have easily derived all the Kinematics equations
The best advantages with calculus method is that we can use for non-uniform acceleration motion also
Watch this tutorial for more information on Uniform Acceleration Motion
How to Solve Kinematics equations problem
Read the question carefully.
Choose the origin of the coordinates and the axes wherever you would like to make calculation easier
Choose one direction as positive . Usually we choose right of the x-axis as positive. The velocity and acceleration in the direction of positive axis will be positive and opposite direction will be negative
Find out the known and unknown. Consider quantities both at the start and end of the time interval.Also it would be good to translate some stated terms in physical quantities such as starting from rest means v_{0}=0
Check out which kinematics equation may be useful in this situation and solve accordingly to get the result.Sometime multiple equation may be required to get the solution
Check for answer If it is reasonable
Be careful about the units used. It should be same
Solved Examples
Question
A boy throws a ball upwards with an initial speed of 20.0 m/s
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player's hands ? (Take g = 10 m/s^{2} and neglect air resistance).
Solution
(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
(b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 10 m/s^{2}
(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
(d) For Upward Motion
Initial velocity of the ball, u = - 20 m/s ( As upward direction and opposite direction to the positive we have choosen)
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a = +g = 9.8 m/s^{2} ( As downward direction and same direction to the positive we have choosen)
From third equation of motion, height (s) can be calculated as:
v^{2} = (v_{0})^{2} + 2a ( x - x_{0} )
Now x=0
0^{2}= (-20)^{2} +2(10)( 0-x_{0})
x_{0}= 20 m
Let t be the time taken to reach the highest point
From First equation of motion
v=u + at
0=-20+ 10t
t= 2sec
As Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the boy hands = 2 + 2 = 4 s.
Quiz Time
Distance travelled in nth sec in Uniform Accelerated Motion
Here is the Summary for Kinematic equations for uniformly accelerated Motion
The above three equations are called the Kinematic equations for uniformly accelerated Motion
$v= v_0 + at$
$x - x_0= v_0t + \frac {1}{2}at^2$
$v^2= v_0^2 + 2a ( x - x_0 )$