the velocity v of a particle moving along the x- axis varies with time t as shown. at t=0 s the particle is at x= -7m.at what time:

• does the particle reach the origin?
  
  
  
   
  
   


• does it return to its starting point at t?
  
  
  
   
  
   


• wot is the distance travelled by yhte particle till its return to the starting pt?
  
  
  
   
  
   

the number written on the positive y axis is 4 , the origin has been marked wid n 'O', the no. on the -ve y axis is -2, the first no. on the +ve x axis is 1 , followed by 3 nd 5, just in case it might become difficult 2 discern dem. 

no i m yet 2 type the complete question

the question's complete now.

sorry cud not upload  the whole image of graph........but here are some important points

at t = 0sec position x = -7

t = 1sec    position x = -3

t= 2sec    position x = 0 (origin)

t = 3 sec  position x = 1

t = 3+sq root2  position x = 0

t = 5   position  x = -1

so here is the answer



particle reach the origin for first time at t = 2 seconds

partice reach the starting point at t = 7 seconds

it travels a distance of 16m till it reach the starting point

PLEASE CORRECT ME IF M WRONG

CASE 1 -  As by graph, its initial velocity is 4m/s towards + x axis so it travels 4m in first second. that means it has to cover 3 m more to reach origin. But after 1 sec, its velocity decreases with constant decelaration, i have said constant because curve after 1 sec is a straight line.

NOW IT TAKES 2 SEC (i.e 1s to 3s) for velocity to become zero. initial velocity was 4m/s and now at time 3 sec it is 0m /s HENCE WE CAN FIND DECELARATION i.e negetive accelaration.

   hence       HENCE ACCELARATION IS - 2 m/s².

NOW WITH THIS ACC, BODY TRAVELS 3 m , LET US SEE IN TWO SEC, HOW MUCH IT TRAVELS ?

 PUTTING u =4m/s, t = 2 sec, and a = -2 m/s²

  WE GET DISPLACE MENT AS 4m HENCE IN TOTAL 3SEC IT TRAVELS 8 METRES.

HENCE IT REACHES THE ORIGIN .

NOW TO FIND AT WHAT TIME IT REACHES ORIGIN, YOU HAVE TO PUT S = 3m AND  FIND THE TIME REST IS SAME. U WILL GET t = 1 SEC. HENCE TOTAL TIME TAKEN FROM START IS 1+1 = 2 S

CASE 2 - YES AS THE VELOCITY BECOMES NEGATIVE, IT WILL AGAIN REACH THE STARTING POINT

CASE 3 - FROM CASE 1 WE HAVE FOUND THAT BODY HAD TRAVELLED MAXIMUM 8 m IN + X AXIS.

HENCE WHEN IT RETURNS, IT TRAVELS SAME 8m SO IT TRAVELS TOTAL OF 16 m.

HOPE THE QUESTION IS SOLVED WITHOUT ANY MISTAKE !