Question : A particle moves a distance x in time t according to equation x=(t+5)-1. The acceleration of particle is proportional to
a) (Velocity)3/2
b) (distance)2
c) (distance)-2
d) (Velocity)2/3
[AIPMT, 2010]
Doubt by Manasvini
Solution :
x=(t+5)-1
we know,
v=dx/dt
v=d[(t+5)-1]/dt
v=-(t+5)-2×d(t+5)/dt
v=-(t+5)-2×(1+0)
v=-(t+5)-2 — (1)
we know,
v=dx/dt
v=d[(t+5)-1]/dt
v=-(t+5)-2×d(t+5)/dt
v=-(t+5)-2×(1+0)
v=-(t+5)-2 — (1)
Now,
a =dv/dt
a = d[-(t+5)-2]/dt
a=-[-2(t+5)-3]×d(t+5)/dt
a=2(t+5)-3×(1+0)
a=2(t+5)-3
Now acceleration can be written as
a=2(t+5)-3×(2/2)
[I have multiplied and divided the power by 3 so that i can include 2 in power.]
a=2(t+5)-2×(3/2)
a=-2[-2(t+5)-2×(3/2)]
a=-2[-2(t+5)-2](3/2)
a=-2[v]3/2
[Using equation (1)]
a∝(v)3/2
a =dv/dt
a = d[-(t+5)-2]/dt
a=-[-2(t+5)-3]×d(t+5)/dt
a=2(t+5)-3×(1+0)
a=2(t+5)-3
Now acceleration can be written as
a=2(t+5)-3×(2/2)
[I have multiplied and divided the power by 3 so that i can include 2 in power.]
a=2(t+5)-2×(3/2)
a=-2[-2(t+5)-2×(3/2)]
a=-2[-2(t+5)-2](3/2)
a=-2[v]3/2
[Using equation (1)]
a∝(v)3/2
a∝(velocity)3/2
Hence, a) (Velocity)3/2, is the correct option.