Question

A body moves from rest with uniform acceleration and travels 270min 3 sec .find the velocity of the body at 10safter the start

Answer 1

here the main thing is uniform acceleration which means that "a" would be constant such that acceleration after 3s would be equal to acceleration after 10 s so from 1st case we will find the acceleration from 2nd equation of motion. i.e s=ut + 1/2 at^{2} from this equation we will find 
   1/2 at2 = s-ut 
  a= 2(s-ut)/t2
  here initial speed u=0 so ut = 0
  a= 2*270/3*3 
  a = 60m/s2 - (1)
  Now we know that v= u+at
from eqn 1 a= 60m/s2
 so here v= 0+ 60*10
 v=600m/s

Answer 2

Hey mate !

_______________________


Given :


Initial velocity (u) = 0 m/s.

Distance (s) = 270 m.

Time (t) = 3 secs.


To find :


Acceleration (a) =?

Velocity after 10 sec.


Solution :

We will use third equation of motion here;

s = ut + 1 / 2 at²

⇒ 270 = 0 * 3 + 1 / 2 * a * 3 * 3

⇒ 270 = 9 / 2 a

⇒ 9a = 540

⇒ a = 540 / 9

⇒ a = 60 m/s²

Now,

We will use first equation of motion here ;

After 10 secs.


v = u + at

⇒ v = 0 + 60 * 10

⇒ v = 600

⇒ v = 600 m/s.


Hence,


The final velocity acquired by the car is 600m/s.


______________________


Thanks for the question !


☺️☺️☺️