Motion in straight Line
6. Motion with constant acceleration
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- 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 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
- 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 v0 at time t=0 plus the change in velocity (at).
- In the same way average velocity can be written as
where x0 is the position of object at time t=0 and vavg is the averag velocity between time t=0 to time t.The above equation then gives
but for the interval t=0 to t the average velocity is
Now from eq. 8 we find
vavg = v0 + ½(at) (11)
putting this in eq. 9 we find
x = x0 + v0t + ½(at2)
x - x0 = v0t + ½(at2) (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,
v2 = (v0)2 + 2a ( x - x0 ) (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 - x0 ) = ½ ( v0 + v ) t (14)
- Same way we can also eliminate v0 using equation 8 and 12. Now from equation 8
v0 = v - at
putting this value of v0 in equation 12 and solving it we finally get,
( x - x0 ) = vt + ½ ( at2 ) (15)
thus equation 15 does not involve initial velocity v0
- Thus these basic equations 8 and 12 , and derived equations 13, 14, and 15 can be used to solve constant acceleration problems.