Motion_Graphs

It represents a change in position of the object with respect to time.

The graph in case the object is stationary (means the distance is constant at all time intervals) – Straight line graph parallel to x = axis

The graph in case of uniform motion – Straight line graph

The graph in case of non-uniform motion – Graph has different shapes

Constant velocity – Straight line graph, velocity is always parallel to the x-axis

Uniform Velocity / Uniform Acceleration – Straight line graph

Non-Uniform Velocity / Non-Uniform Acceleration – Graph can have different shapes

The area under the graph gives the distance traveled between a certain interval of time. Hence, if we want to find out the distance traveled between time interval t1 and t2, we need to calculate the area enclosed by the rectangle ABCD where area (ABCD) = AB * AC.

Similarly, to calculate distance traveled in a time interval in case of uniform acceleration, we need to find out the area under the graph, as shown in the figure below.

To calculate the distance between time intervals t1 and t2 we need to find out area represented by ABED.

Area of ABED = Area of the rectangle ABCD + Area of the triangle ADE = AB × BC + 1/ 2 * (AD × DE)

The line segment PN shows the relation between velocity and time. 

Initial velocity, u can be derived from velocity at point P or by the line segment OP

Final velocity, v can be derived from velocity at point N or by the line segment NR

Also, NQ = NR – PO = v – u

Time interval, t is represented by OR, where OR = PQ = MN

Acceleration = Change in velocity / time taken

Acceleration = (final velocity – initial velocity) / time

a = (v – u)/t

so, at = v – u

v = u + at

We know that, distance travelled by an object = Area under the graph

So, Distance travelled = Area of OPNR = Area of rectangle OPQR + Area of triangle PQN

s = (OP * OR) + (PQ * QN) / 2

s = (u * t) + (t * (v – u) / 2)

s = ut + 1/2 at²    [because at = v – u]

3. Deriving the Equation for Position – Velocity Relation

We know that, distance travelled by an object = area under the graph

So, s = Area of OPNR = (Sum of parallel sides * height) / 2

 s = ((PO + NR)* PQ)/ 2

s = ( (v+u) * t)/ 2

2s / (v+u) = t [equation 1]

Also, we know that, (v – u)/ a = t [equation 2]

On equating equations 1 and 2, we get,

2s / (v + u) = (v – u)/ a

2as = (v + u) (v – u) 

2 a s = v² – u²

If an object moves in a constant velocity along a circular path, the change in velocity occurs due to the change in direction. Therefore, this is an accelerated motion. Consider the figure given below and observe how directions of an object vary at different locations on a circular path.