At time t = 0, a 2150-kg rocket in outer space fires an engine that exerts an increasing force on it in the + x-direction. This force obeys the equation F = At^2, where t is time,

and has a magnitude of 781.25 N when r = 1.25 s. (a) Find the SI value of the constant A, including its units. (b) What impulse does the engine exert on the rocket during the I. 50-s

interval starting 2.00 s after the engine is fired? (c) By how much does the rocket? s velocity change during this interval?

a) A = 500 N/s²

b) I = 5812.50 N-s

c) Δv = 2.7034 m/s

Explanation:

Given info

m = 2150 kg

F (t) = At²

F (1.25 s) = 781.25 N

a) A = ?

We use the equation

F (t) = At² ⇒ 781.25 N = A * (1.25 s) ²

⇒ A = F (t) / t² = (781.25 N) / (1.25 s) ²

⇒ A = 500 N/s²

b) I = ? if 2.00 s ≤ t ≤ 3.50 s

we apply the equation

I = ∫F (t) dt = ∫At² dt = A ∫t² dt = (500/3) * t³ + C

Since the limits of integration are 2 and 3.5, we obtain

I = (500/3) * ((3.5) ³ - (2) ³) = 5812.50 N-s

c) Δv = ?

we can apply the equation

I = m*Δv ⇒ Δv = I / m

⇒ Δv = 5812.50 N-s / 2150 kg

⇒ Δv = 2.7034 m/s