Question

A particle moves along a path x = y²+2y+2 plane such that the y-component of its velocity remains constant at 2 m/s. The magnitude of its acceleration is​

Answer 1

Explanation:

ANSWER

x=y

2

+2y+2

v

x

=

dt

dx

=2y

dt

dy

+2

dt

dy

=2v

y

(y+1)

since the y component of velocity remains the same ,there is no acceleration along the y component,a

y

=0 a

x

=

dt

dv

x

=2v

y

(

dt

dy

)=2v

y

v

y

=2×5×5=50ms

−2

.