KEAM 2013 Physics and Chemistry Paper

Given, x2=2t2+6t+1....(i)
Differentiating Eq. (i) w.r.t. t, we get
2x‌

dx

dt

=4t+t
2xv=4t+6  (∵v=‌

dx

dt

)
xv=2t+3......(ii)
Now, again differentiating Eq. (ii) w.r.t. t, we get
x‌

dv

dt

+v‌

dx

dt

=2
x⋅a+v⋅v=2‌‌(∵a=‌

dv

dt

and v=‌

dx

dt

)
Here, v2=‌

4t2+12t+9

x2

v2=‌

2(2t2+6t+1)+7

x2

v2=‌

2t2+7

x2

.....(iv)
Put the value of v2 in Eq. (iii) from Eq. (iv), we get
xa+‌

2x2+7

x2

=2
x3a+2x2+7=2x2
x3a+7=0
x3a=−7
a=‌

−7

x3

Hence, the acceleration varies with x−3.