Question

A particle moves along a path x = y^2 + 2y + 2 in a plane such that the y-component of its velocity remains constant at 2 m/s. The magnitude of its acceleration is?

Answer 1

Answer:

Since velocity in y is constant, thus acceleration in y=0. Thus magnitude of acceleration is given by the x-component of acceleration.

Now, x=y

2

+2y+2.

Differentiate with time to get,

v

x

=2yv

y

+2v

y

=10y+10

Again differentiate with time to get,

a

x

=10v

y

Since v

y

=5, thus a

x

=50m/s

2