AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Given, the expression of the velocity is,
v=x

^

i

+yt

^

j

The magnitude of the velocity is,
|v|=√x2+y2t2
The magnitude of the tangential acceleration is,
at=

d|v|

dt

at=

d√x2+y2t2

dt

at=

ty2

√x2+y2t2

The tangential acceleration at t=

x√3

y

is,
at=

( √3 y )y2

√x2+y2( x√3 y )2

=

√3y

2

The total acceleration is,
a=

dv

dt

a=

d

dt

(x

^

i

+yt

^

j

)
a=y

^

j

The magnitude of the normal acceleration is,
|an|=√a2−ab2
=√y2−(

y4t2

x2+y2t2

)
=√

x2y2

x2+y2t2

=

xy

√x2+y2t2

The magnitude of normal acceleration at t=

x√3

y

is,
|an|=

xy

√x2+y2 3x2 y2

=

y

2