Question

Starting from rest, the boy runs outward in the radial direction from the center of the platform with a constant acceleration of 0.5 m/s2 . The platform rotates at a constant rate of 0.2 rad/s. Determine his velocity (magnitude) when t = 3 sec

Answer 1

The boy's tangential velocity, after accelerating radially for 3 seconds at 0.5 m/s^2 while the platform rotates at 0.2 rad/s, is 1.5 m/s.

Step-by-step explanation:

To determine the boy's velocity (magnitude) at t = 3 sec while he runs outward radially from the center of a rotating platform, we must consider both his radial (tangential) velocity due to his acceleration and the rotational movement of the platform.

The boy starts from rest with a constant radial acceleration of 0.5 m/s2. Using the kinematic equation v = u + at (where u is the initial velocity, a is the acceleration, and t is the time), we can calculate his radial velocity after 3 seconds as:

vradial = 0 + (0.5 m/s2)(3 s) = 1.5 m/s

The platform's constant rotation rate is given as 0.2 rad/s. This rotational movement does not change the boy's radial (tangential) velocity directly, as the question only asks for his velocity in the radial direction. Therefore, the total magnitude of the boy's velocity after 3 seconds is 1.5 m/s tangentially.

Answer 2

His velocity when t= 3 sec is V= 1.56 m/s

Step-by-step explanation:

a= 0.5 m/s²

ω= 0.2 rad/s

t= 3 sec

Vr= a*t

Vr= 1.5 m/s

r= a*t²/2

r= 2.25m

Vt= w*r

Vt= 0.45 m/s

V= √(Vr²+Vt²)

V= 1.56 m/s