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 Fx = At2, where t is time, and has a magnitude of 781.25 N when t = 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 1.50 s interval starting 2.00 s after the engine is fired? c) By how much does the rockets velocity change during this interval?

Answer 1

Answer:

The SI unit of A = N/s^2

Explanation:

Given the equation:

Fx = At^2

Where Fx = magnitude of force = 781.25N

t = 1.25 seconds

Rearranging the equation to make A subject of formular gives:

A = Fx/t^2

A = 781.25/(1.25^2)

A = 781.25/1.5625

A = 500N/s^2

Therefore, SI unit of A = N/s^2

b) Impluse,I = Force × time

Impluse ,I = Integral( f × dt) at intervals of t2 =2.0s and t2 = 1.5s

Impluse, I = f (t2) - f(t1)

Impluse ,I = 781.25 (2.0) - 781.25(1.5)

Impluse,I = 1562.5 - 1171.88

Impluse = 390.63Ns

c) The change in Velocity during the interval can be determined by the given equation:

Impluse ,I = M × change in velocity

Where M = mass

But mass, m = f/g

M = 781.25/9.8

M = 79.72kg

I = M × change in velocity

Rearranging the equation gives:

Change in velocity = Impluse/ mass

Change in velocity = 390.63/ 79.72

Change in velocity = 4.9m/s

Answer 2

Answer:

(a) A = 499.84N/s²

(b) Impulse = 771Ns

(c) Δv = 0.359m/s

Explanation:

Please see attachment below.

To find the impulse in part (b) we integrate the the function for force with respect to time since it is time dependent between the given time intervals.