# Graph between the square of velocity (v) of a particle and the distance (s) moved is shown in figure. The acceleration of the particle in kilometers per hour square is:

By using the 3rd equation of Velocity –displacement equation -

$V^{2}-u^{2}=2as$

$V\rightarrow$ Final Velocity = 4600

$s\rightarrow$ Displacement

$u\rightarrow$Initial velocity = 950

$a\rightarrow$acceleration

-

We use V2 = u2 + 2as

V2 - u2 = 2as
By putting the values in above equation and solving we get
a = 3041.6

### Preparation Products

##### Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
##### Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
##### Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-