Question

A 2.0 kg projectile with initial velocity v⃗ = 6.5 ı^ m/s experiences the variable force F⃗ = -2.0tı^ + 4.0t2ȷ^ N, where t is in s. a. What is the projectile's speed at t = 2.0 s? b. At what instant of time is the projectile moving parallel to the y-axis?

Answer 1

Answer:

a. |v| = 6.98 m/s

b. t = 3.61 s

Explanation:

Acceleration = force / mass

a = F / m

a = (-2.0t i + 4.0t² j) / (2.0)

a = -1.0t i + 2.0t² j

Integrate to get velocity.

v = (-½ t² + C₁) i + (⅔ t³ + C₂) j

At t = 0, v = 6.5 i.

C₁ = 6.5

C₂ = 0

v = (-½ t² + 6.5) i + (⅔ t³) j

a. At t = 2.0, the velocity is:

v = (-½ (2.0)² + 6.5) i + (⅔ (2.0)³) j

v = 4.5 i + 5.33 j

Speed is the magnitude of velocity.

|v| = √(4.5² + 5.33²)

|v| = 6.98 m/s

b. The time when the x component of the velocity is 0 is:

0 = -½ t² + 6.5

½ t² = 6.5

t² = 13

t = 3.61 s