A 0.50kg object is at rest. a 3.00N force to the right acts onthe object during a time interval of 1.50s.

a. what is the velocity of the object at the end of this timeinterval?

b. at the end of this interval, a constant force of 4.00N tothe left is applied for 3.00s. what is the velocity at the end ofthe 3.00s?

(a) 6.0 m/s to the right

(b) 24.0 m/s to the left

Explanation:

Velocity: This can be defined as the time rate of change of displacement of a body. Velocity is a vector quantity, and as such it can be represented both in magnitude and direction.

From the question,

F = ma (Newton's fundamental equation of dynamics)

a = F/m ... Equation 1

Where F = force, m = mass of the object, a = acceleration.

v = u + at ... Equation 2.

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

(a) Given: F = 3.00 N, m = 0.50 kg.

Substituting into equation 1,

a = 3/0.5

a = 6.0 m/s².

Also Given: t = 1.5 s, u = 0 (at rest)

Substituting into equation 2.

v = 0 + 6 (1.5)

v = 9 m/s to the right.

(b) Given: F = 4.00 N, m = 0.50 kg.

Substituting into equation 1

a = 4/0.5

a = 8 m/s²

Also given: t = 3.00 s, u = 0 m/s (at rest)

Substituting into equation 2

v = 0 + 3 (8)

v = 24 m/s to the left.