Question

Give the 3 equations of motion and define the terms​

Answer 1

Explanation:

1)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.

It has been derived here by the graphical method.

2)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.

3)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.

Hope it's helpful for you!

Answer 2

Answer:

Explanation:

1)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.

It has been derived here by the graphical method.

2)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.

3)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.

Hope it's helpful for you!