AIPMT 2010 Physics and Chemistry Paper

Distance, x=(t+5)−1......(i)
Velocity, v=‌

dx

dt

=‌

d

dt

(t+5)−1
=−(t+5)−2 .....(ii)
Acceleration, a=‌

dv

dt

=‌

d

dt

[−(t+5)−2]
=2(t+5)−3 ...(iii)
From equation (ii), we get
v3/2=−(t+5)−3....(iv)
Substituting this in equation (iii) we get
Acceleration, a=−2v3/2
or a ∝(‌ velocity ‌)3/2
From equation (i), we get
x3=(t+5)−3
Substituting this in equation (iii), we get
Acceleration, a=2x3
or a∝(‌ distance ‌)3