a problem in kinematics

from graph we get eq

a=1-t

now a=d²xdt²

ab karo.

hmmm..........
so after double integration i get s = t²/2 - t³/6.......
still not gettin the correct ans!!

observe from graph that eq. of the given line is of form→ y= 1-x
which means that a_y=1-a_x

a_y+ a_x =1

dydt + dxdt =1

dy + dx = dt

let dy +dx = dr ( in vector form)

so, dr = dt

now integrating ....
\large \int_{0}^{r}{dr}= \int_{0}^{4}{dt}

which gives r = 4 m.

@akhil
here distance is asked not displacemnt

here v=t- (t²/2)

so distnce travelled is =∫|v| dt

thnx bornidentity!!!!

above sol. that i posed is wrong.

from graph.
a= 1-t

dvdt=1-t

integrating the above eq. ( 0→v and 0→t ) we get→

v = t - t²2 = t(1-t2)

since we need distance we have to take mod of velocity.

v= t(1-t2) , 0≤t≤2

t(t2-1) , 2 ≤t≤4

integrating 1st function within required limit we get, x = 23

similarily integrating 2nd function within required limit we get, x= 103

total distance = 23 + 103 = 4 m

now u got it rite!!!!!!!:D

lol :)