Question

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) = At + Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.40 m/s^2 and 1.80 s later an upward velocity of 2.18 m/s .


A) Determine A.
A=..1.40..m/s^2

B) Determine B.
B=......M/s^3

C) At 5.00s after fuel ignition, what is the acceleration of the rocket?
a=....m/s^2

D) What thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust in newtons.
T=......N

E) What thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust as a multiple of the rocket's weight.
T=.....w

Answer 1

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

Answer 2

Part A: The value of A is 1.4 m/s^2.

Part B: The value of B is -0.104 m/s^2.

Part C: The acceleration of the rocket at t = 5s is 0.36 m/s^2.

Part D: The thrust force of the rocket is 25704.8 N.

Part E: The thrust force as multiple of rocket's weight is 1.03 W.

What is acceleration?

Acceleration is defined as the rate at which velocity changes with time, in terms of both speed and direction.

Given that the mass m of the rocket is 2530 kg and its velocity function is v(t) = At + Bt^2. At the instant of ignition, the rocket has an upward acceleration of 1.40 m/s^2 and 1.80 s later an upward velocity of 2.18 m/s.

The acceleration of the rocket is given below.

$a = \dfrac {dv}{dt}$

$1.4 = \dfrac {d (At +Bt^2}{dt}$

$1.4 = A + 2Bt$

Part A:

Lets put t = 0, then

$1.4 = A + 2B \times 0$

$A = 1.4\;\rm m/s^2$

The value of A is 1.4 m/s^2.

Part B:

The velocity at t = 1.80 is 2.18 m/s. So,

$v = At + Bt^2$

$B = \dfrac {v-At}{t^2}$

$B = \dfrac {2.18 - 1.4\times 1.80}{1.80^2}$

$B = -0.104 \;\rm m/s^2$

The value of B is -0.104 m/s^2.

Part C:

At t = 5 s, the value of acceleration is given below.

$a = A + 2Bt$

$a = 1.4 + 2 \times (-0.104) \times 5$

$a = 0.36 \;\rm m/s^2$

The acceleration of the rocket at t = 5s is 0.36 m/s^2.

Part D:

The thrust force of the rocket is given below.

$T = m(a+g)$

Where g is the gravitational acceleration and a is the acceleration of the rocket.

$T = 2530 ( 0.36 + 9.8)$

$T = 25704.8 \;\rm N$

The thrust force of the rocket is 25704.8 N.

Part E:

The weight of the rocket is given below.

$W = m\times g$

$W = 2530 \times 9.8$

$W=24794$

The final thrust of the rocket is given below.

$T' = \dfrac {T}{W}$

$T' = \dfrac { 25740.8}{24794}$

$T' = 1.03\;\rm W$

Hence we can conclude that the thrust force as multiple of rocket's weight is 1.03 W.

To know more about the acceleration, follow the link given below.

https://brainly.com/question/2437624.