Slide 1 : KINEMATICS
Graphs © 2007-08 P K Bharti, IIT Kharagpur www.vidyadrishti.com
Slide 2 : P. K. BHARTI, I.I.T. KHARAGPUR From the velocity (along x axis) vs. time plot shown, find (a)the distance travelled by the particle during first 40s, (b) the avg. velocity during this period, (c) avg. acceleration between 0 to 10s, and (d) acceleration at t=5s.
SOLUTION: t(s) v (m/s) 0 20 40 5 -5
Slide 3 : P. K. BHARTI, I.I.T. KHARAGPUR From theory we know that, from area of velocity time graph we get change in position, i.e., displacement. We also know that area above t-axis is positive and below t-axis is negative.
Therefore, displacement between t=0s to t=20s
= area of v-t graph = area of triangle = ½ x 20 x 5 =50m.
And, displacement between t=20s and t=40s
= - (area of v-t graph) [-ve, because area below t-axis]
= - ½ x 5 x 20 = -50m.
(a) Distance is the total length of the path.
Therefore distance travelled by the particle during the first 40s
= 50 + 50 =100m (ans)
CONTINUED….
Slide 4 : P. K. BHARTI, I.I.T. KHARAGPUR (b). Avg. velocity during first 40s
= (change in position during this period)/(time change)
=(total displacement during this period)/(total time taken)
=(50-50)/(40-0) = 0. (ans)
(c). Avg. acceleration 0 to 10s
= (change in velocity)/(time taken)
= (5-0)/(10-0) = 0.5 m/s2. (ans)
CONTINUED...
Slide 5 : P. K. BHARTI, I.I.T. KHARAGPUR (d). We have to find acceleration at t=5s, i.e., we have to find instantaneous acceleration (because at a required acceleration is at particular instant of time) at t=5s.
Instantaneous acceleration is found by slope of tangent at that instant.
Therefore,
a = tan?
= p/b
=5/10
= 0.5 m/s2.
(ans) t(s) v (m/s) 0 20 40 5 -5 5 . ? 10
Conversion of graphs : P. K. BHARTI, I.I.T. KHARAGPUR Conversion of graphs Sometimes u r asked to convert velocity-time graph to accln vs. time graph or distance vs. time graph etc.
For these types of graph simply differentiation and integration principal holds.
Find out the equation of the given graph (if possible) and then differentiate or integrate [according to question] to find the required graph.
Coming slides explain the conversion of some standard graphs.
Slide 7 : P. K. BHARTI, I.I.T. KHARAGPUR x = at2+bt+c [parabola]
x’ = 2at + b [line]
x” = 2a
[line parallel to t-axis]
* x’” = 0 [t-axis]
E.g., suppose velocity vs. time graph is a line, then accln vs. time graph will be a line parallel to t-axis and distance vs. time graph will be a parabola.
NOTE: x’, x” and x”’ represents derivative, double derivative and triple derivative of x with respect to t respectively. Differentiation Integration
Slide 8 : P. K. BHARTI, I.I.T. KHARAGPUR SOME EXAMPLES FOR SHORTCUT: motion Time, t Position (linear& slope is positive) Velocity (constant +ve) Acceleration (zero) o
Slide 9 : P. K. BHARTI, I.I.T. KHARAGPUR motion Time, t Position (linear & slope is negative) Velocity (constant -ve) Acceleration (zero) o
Slide 10 : P. K. BHARTI, I.I.T. KHARAGPUR motion Time, t Position (upward parabola
[one half] ) Velocity (linear) Acceleration (constant) o
Conversion of standard graphs : P. K. BHARTI, I.I.T. KHARAGPUR Conversion of standard graphs motion Time, t Position (upward parabola [one half]) Velocity (linear) Acceleration (constant) o
Conversion of standard graphs : P. K. BHARTI, I.I.T. KHARAGPUR In short for upward parabola position vs. time graph we have Conversion of standard graphs motion Time, t Position (upward parabola) Velocity (linear) Acceleration (constant) o
Slide 13 : P. K. BHARTI, I.I.T. KHARAGPUR motion Time, t Position (downward parabola
[one half] ) Velocity (linear) Acceleration (constant) o
Slide 14 : P. K. BHARTI, I.I.T. KHARAGPUR motion Time, t Position (downward parabola
[one half]) Velocity (linear) Acceleration (constant) o
Slide 15 : P. K. BHARTI, I.I.T. KHARAGPUR In short for downward parabola position vs. time graph we have: motion Time, t Position (downward parabola) Velocity (linear) Acceleration (constant) o
Slide 16 : P. K. BHARTI, I.I.T. KHARAGPUR A particle travels along the x-axis whose velocity (along x-axis) vs. t plot is shown. Assuming the coordinate of the particle x=0 at the moment t=0, draw the approximate time dependence plots of acceleration a, the x coordinate (i.e., position) and the distance covered s. -2 t v 8 O A B C D E Example
Slide 17 : P. K. BHARTI, I.I.T. KHARAGPUR -2 t A B D E v a x s Example
Slide 18 : P. K. BHARTI, I.I.T. KHARAGPUR LDETAILS:
Let us first write eqns. of all 5 line segments using eqn. (From coordinate geometry)
Replacing y by v and x by t , we get eqns. as:
Line segment OA Example
Slide 19 : P. K. BHARTI, I.I.T. KHARAGPUR Similarly, eqns. of
Line segment AB: v = 1 …(2)
Line segment BC: v = -t+4 …(3)
Line segment CD: v = 2t-14 …(4)
Line segment DE: v=0 …(5)
Now, for acceleration we have to differentiate v wrt t and for position we have to use integration.
Distance is nothing but total length of the journey, so it will be either equal to +ve displacement or equal to –(–ve dispalcement).
CONTINUED… Example
Slide 20 : P. K. BHARTI, I.I.T. KHARAGPUR For time t=0 to t=1
*v = t ? a = dv/dt =1 . Therefore, a=1
It means for time t=0 to t=1 accln remains constant to 1.
*Again, v = t ? dx/dt =t? ? dx = ?t dt ? x = t2/2 + c1
where c1 is a constant.
Given at t=0, x=0 ? x = t2/2+c1 ? 0=0+c1 ? c1 = 0
Therefore, x=t2/2 (Upward parabola). Clearly, @t=1 , x= ½
* Since, displacement is positive
? Distance (s) = displacement. t A O a x v (1, ½ ) s Example
Slide 21 : P. K. BHARTI, I.I.T. KHARAGPUR For time t=1 to t=3
*v = 1 ? a = dv/dt = o ? a=0
It means for time t=1 to t=3 accln remains constant to 0.
*Again, v=1 ? dx/dt =1? ? dx = ?dt ? x = t+ c2
where c2 is a constant. Now from previous part, @t=1, x= ½
? x=t+c2 ? ½ = 1 + c2 ? c2 = -½ .
Therefore, x = t - ½ . So @ t = 3 , x = 3-½ ? x = 5/2
*Again displacement is positive, so s = x. (3, 5/2) (1, ½ ) v a x s Example
Slide 22 : P. K. BHARTI, I.I.T. KHARAGPUR Follow the same procedure and get graphs for another parts.
You will find, after t position x starts decreasing, so displacement is negative, it means distance after (t=4) = -(displacement). Use this concept to draw s vs. t graph after t=4. Example