Question

derive three equations of motion using distance time and distance velocity graph
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Answer 1

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• derive three equations of motion using distance time and distance velocity graph ..

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The three equations of motion v = u + at; s = ut + (1/2) at2 and v2 = u2 + 2as can be derived with the help of graphs as described below.

1. Derive v = u + at by Graphical Method

Consider the velocity – time graph of a body shown in the below Figure.

Velocity – Time graph to derive the equations of motion.

Velocity-Time graph to derive the equations OOF motion

The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE

Now, Initial velocity of the body, u = OA ------- (1)

And, Final velocity of the body, v = BC -------- (2)

But from the graph BC = BD + DC

Therefore, v = BD + DC -------- (3)

Again DC = OA

So, v = BD + OA

• Now, From equation (1), OA = u
• So, v = BD + u --------- (4)
• We should find out the value of BD now.

We know that the slope of a velocity – time graph is equal to acceleration, a

Thus, Acceleration, a = slope of line AB

or a = BD/AD

But AD = OC = t,

so putting t in place of AD in the above relation, we get:

• a = BD/t

or BD = at

Now, putting this value of BD in equation (4) we get

• v = at + u

This equation can be rearranged to give:

v = u + at

And this is the first equation of motion

2. Derive s = ut + (1/2) at2 by Graphical Method

Velocity-Time graph to derive the second equation of motion

Velocity–Time graph to derive the equations of motion.

Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus

Distance travelled = Area of figure OABC

= Area of rectangle OADC + Area of triangle ABD

We will now find out the area of the rectangle OADC and the area of the triangle ABD.

(i) Area of rectangle OADC = OA × OC

= u × t

= ut ...... (5)

(ii) Area of triangle ABD = (1/2) × Area of rectangle AEBD

= (1/2) × AD × BD

= (1/2) × t × at (because AD = t and BD = at)

= (1/2) at2 ------ (6)

So, Distance travelled, s = Area of rectangle OADC + Area of triangle ABD

or s = ut + (1/2) at2

This is the second equation of motion. It has been derived here by the graphical method.

Derive v2 = u2 + 2as by Graphical Method

Velocity-Time graph to derive the third equation of motion

Velocity–Time graph to derive the equations of motion.

We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

In other words,

Distance travelled, s = Area of trapezium OABC

Distance travelled =Area of trapezium

Now, OA + CB = u + v and OC = t.

Putting these values in the above relation, we get

------- (7)

We now want to eliminate t from the above equation.

This can be done by obtaining the value of t from the first equation of motion.

Thus, v = u + at (First equation of motion)

And, at = v – u or

Now, putting this value of t in equation (7) above, we get:

or 2as = v2 – u2 [because (v + u) × (v – u) = v2 – u2]

or v2 = u2 + 2as

This is the third equation of motion

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ANOTHER SMALL.ANSWER=>Deriving 3 equations of motion (from v-t graph) Let's derive the three equations of motion using a velocity time graph v = u + at s = ut + 1/2 at^2 v^2 = u^2+2as

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Hope this help