Question

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Answer 1

Answer:

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Let's break the equation into pieces.. On L.H.S.:

Let's break the equation into pieces.. On L.H.S.:V2V2is scalar.

U2U2 is scalar

So, right hand side should be also a scalar.

Now a is vector. So, S must be a vector and it should be a.S (dot product)

Wait! But why do we write aS instead of a.S ?

Here, a is uniform acceleration. That is acceleration doesn't change in magnitude nor in direction. This point is very important.

The object can never change its direction.

Thus it is a straight line motion and in a straight line motion, a and S are aligned or a.S is simply aS

Also, note that even though S is a displacement vector, its magnitude should be equal to distance scalar only.

Thus, for time being without any harm, one can replace S with d.

Note: I'm not saying you can replace S with d.

Explanation:

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Answer 2

1. Derive v2 = u2 + 2as by Graphical Method



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 

 

 

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