Question

02. The motion of a particle in the xy-plane is given by x(t) = 25+6t? m; y(t) = -50-20t+8t2 m. The magnitude of the initial velocity of the particle. Vo is given by 30 m s 1) 2) 40 m s 3) 50 m st 20 m s

Answer 1

Answer:

6897

Explanation:

vehhhu is the 7th ijgddgbuy

Answer 2

Therefore the magnitude of the initial velocity of the particle is 20 m/s.( Option - 4 )

Given:

The particle is moving in the XY-plane and the motion is given by

x(t) = 25+6t

y(t) = -50-20t+8t²

To Find:

The magnitude of the initial velocity of the particle at time t = 0.

Solution:

This numerical can be easily solved by the following process.

Given displacement equations,

⇒ x(t) = 25+6t

⇒ y(t) = -50-20t+8t²

The first derivative of the displacement equation gives the Velocity equation.

⇒ Vₓ = d (x(t)) / dt = d ( 25 + 6t )/dt = 6

⇒ Vy = d (y(t)) / dt = d ( -50-20t+8t² ) /dt = -20 + 16t

At time t = 0

⇒ Vₓ = 6

⇒ Vy = -20 + 0 = -20

Total Velocity = √ Vₓ² + Vy²  = √ 6² + (-20)² = √ 36 + 400 = √ 436 ≈ 20

Therefore the magnitude of the initial velocity of the particle is 20 m/s.

#SPJ3