Advertisements
Advertisements
Question
Derive the following equation of motion by the graphical method : v2 = u2 + 2as, where the symbols have their usual meanings.
Advertisements
Solution
In the given figure, the distance travelled (s) by a body in time (t) is given by the area of the figure OABC which is a trapezium.
Distance travelld = Area of the trapezium OABC
So, Area of trapezium OABC,
= `"(Sum of parallel sides)(Height)"/2`
=`"(OA+CB)(OC)"/2`
Now, (OA + CB) = u + v and (OC) = t.
Putting these values in the above relation, we get:
`s = ((u+v)/2)t` ....(1)
Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at
So,
`t = "v-u"/a`
Now, put this value of t in equation (1), we get:
`s = (((u+v)(v-u))/(2a))`
On further simplification,
2as = v2 – u2
Finally the third equation of motion.
`v^2 = u^2 + 2as`
where
(s) - Displacement
(u) - Initial velocity
(a) - Acceleration
(v) - Final velocity
(t) - Time taken
APPEARS IN
RELATED QUESTIONS
Name the two quantities, the slope of whose graph give acceleration.
Draw a velocity-time graph to show the following motion :
A car accelerates uniformly from rest for 5 s ; then it travels at a steady’ velocity for 5 s.
Define speed. What is its S.I. unit?
Multiple choice Question. Select the correct option.
A graph is a straight line parallel to the time axis in a distance-time graph. From the graph, it implies:
A train starting from rest picks up a speed of 20 ms−1 in 200 s. It continues to move at the same rate for the next 500 s and is then brought to rest in another 100 s.
- Plot a speed-time graph.
- From graph calculate
(a) uniform rate of acceleration
(b) uniform rate of retardation
(c) total distance covered before stopping
(d) average speed.
From the diagram given below, calculate distance covered by body.
Represent the position of a body described as at 15 m, 30o north of west, on a graph paper.
Name the physical quantity which is equal to the area under speed-time graph.
Draw the following graph:
Speed versus time for a non-uniform acceleration.
(i) In speed time graph uniform motion is given by a straight line parallel to x axis so figure (a) denotes the uniform motion.
(ii) In speed time graph motion with increasing speed is shown by straight line with positive slope so figure (c) denotes the motion with speed increasing.
(iii) In speed time graph motion with decreasing speed is shown by straight line with negative slope so figure (b) denotes the motion with speed decreasing.
(iv) In speed time graph motion with oscillating speed is shown by zigzag line so figure {d) denotes the motion with speed oscillating.
