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
The value of constant A can be calculated by substituting the value of F and t. The impulse can be calculate by integration of given equation of f(t).
(a) The SI value of the constant A, including its units is [tex]500 \:\rm N/s^2[/tex].
(b) The impulse exerted on rocket is [tex]5812.50 \:\rm N-s[/tex].
(c) The change in velocity is [tex]2.7034 \: \rm m/s[/tex].
Given:
The rocket mass is [tex]m=2150\:\rm kg[/tex].
The time is [tex]t=1.25\: \rm s[/tex]
The magnitude is [tex]F=781.25\:\rm N[/tex]
(a)
The force obeys the equation as follows,
[tex]F = At^2[/tex]
Substitute the value.
[tex]781.25 \:\rm N =A\times(1.25 \:\rm s)^2\\A=500 \:\rm N/s^2[/tex]
(b)
The impulse is defined as the integral of the force.
Calculate the impulse exerted on rocket.
[tex]I=\int F(t)\\I=\int At^2 dt\\I=A \int t^2dt\\I=\dfrac{500}{3}\times t^3+ C[/tex]
Substitute the integration limit from 2 and 3.5 s.
[tex]I=\dfrac{500}{3}\times ((3.5)^3-(2)^3)\\I= 5812.50 \: \rm N-s[/tex]
(c)
Calculate the change in velocity.
[tex]I = m\times \Delta v\\\Delta v =\dfrac{I}{m}[/tex]
Substitute the value.
[tex]\Delta v=\dfrac{5812.50\:\rm N-s}{2150 \: \rm kg}\\\Delta v=2.7034\:\rm m/s[/tex]
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