Question

Topic : Solving Kinematic equation by graphical method.

1. Derive by graphical method : v = u + at

2. Derive by graphical method : S = ut + ½ at2

3. The figure given on the next page shows distance time graphs

Of three objects A, B and C. Study the graphs and answer the

following questions.

a) Which of the three is travelling the fastest?

b) Are all three ever at the same point on the road ?

c) How far has C travelled when B passes A?

d) How far has B travelled by the times it passes C?

plz answer fast I need it
I will mark as brilliant answer.​

Answer 1

Answer:

Let the initial velocity of the object = u

Let the object is moving with uniform acceleration, a.

Let object reaches at point B after time, t and its final velocity becomes, v

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to y-axis which meets at E at y-axis.

Let OE = time, t

Now, from the graph,

BE = AB + AE

⇒ v = DC + OD (Since, AB = DC and AE = OD)

⇒ v = DC + u (Since, OD = u)

⇒ v = DC + u ------------------- (i)

Now, Acceleration (a) =Change in velocityTime taken=Change in velocityTime taken

⇒a=v−ut⇒a=v-ut

⇒a=OC−ODt=DCt⇒a=OC-ODt=DCt

⇒at=DC⇒at=DC -----(ii)

By substituting the value of DC from (ii) in (i) we get

v=at+uv=at+u

⇒v=u+at⇒v=u+at

Above equation is the relation among initial vlocity (uu), final velocity (vv), acceleration (a) and time (t). It is called first equation of motion.

Equation for distance –time relation

Distance covered by the object in the given time ‘t’ is given by the area of the trapezium ABDOE

Let in the given time, t the distance covered by the moving object = s

The area of trapezium, ABDOE

= Distance (s) = Area of △ABD+△ABD+ Area of ADOE

⇒s=12×AB×AD+(OD×OE)⇒s=12×AB×AD+(OD×OE)

⇒s=12×DC×AD+(u+t)⇒s=12×DC×AD+(u+t)

[Since, AB=DCAB=DC]

⇒s=12×at×t+ut⇒s=12×at×t+ut

⇒s=12×at×t+ut⇒s=12×at×t+ut

[∵ DC=atDC=at]

⇒s=12at2+ut⇒s=12at2+ut

Answer 2

Answer:

Sorry I don't know this question