The movement of a material point is described by the equation x = 5 -2t + t ^2. Taking its mass equal to 1.5 kg

The movement of a material point is described by the equation x = 5 -2t + t ^2. Taking its mass equal to 1.5 kg, find the impulse in 3 s and 5 s after the start of the movement and the change in impulse for the specified time.

x (t) = 5 – 2 * t + t ^ 2.

m = 1.5 kg.

t1 = 3 s.

t2 = 5 s.

p1 -?

p2 -?

Δp -?

The momentum of a body p is a vector physical quantity equal to the product of the body’s mass m by its speed of movement V: p = m * V.

The dependence of the body’s velocity V (t) will be the derivative of the dependence of the x (t) coordinate “: V (t) = x (t)”.

V (t) = (5 – 2 * t + t ^ 2) “= -2 + 2 * t.

V1 (t1) = -2 + 2 * 3 = 4 m / s.

p1 = 1.5 kg * 4 m / s = 6 kg * m / s.

V2 (t2) = -2 + 2 * 5 = 8 m / s.

p2 = 1.5 kg * 8 m / s = 12 kg * m / s.

Δp = p2 – p1.

Δp = 12 kg * m / s – 6 kg * m / s = 6 kg * m / s.

Answer: p1 = 6 kg * m / s, p2 = 12 kg * m / s, Δp = 6 kg * m / s.