JEE Main 16 March 2021 Shift 1 Solved Paper

From the given velocity-displacement graph, for 0≤x≤200, the graph is a straight line i.e., varying linearly.
As, the equation of straight line is
y=mx+c
Here, it can be given as
v=mx+c
where, m= slope =‌

(50−10)

200−0

=‌

1

5

and c=10
⇒‌‌v=‌

x

5

+10
As, acceleration, a=‌

dv

dt

=‌

vdv

dx

a=(‌

x

5

+10)‌

d

dx

(‌

x

5

+10)  [using Eq. (i)]
=(‌

x

5

+10)(‌

1

5

)=2+‌

x

25

For x=0,a=2+‌

0

25

=2ms−2
and for x=200,a=2+‌

200

25

=10ms−2
Therefore, the graph will have straight line till x=200 and for x>200;v= constant and a=0
So, from the graphs given in the options, the above conditions are satisfied in option (a) only.