Question

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?

Answer 1

Answer:

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