Question

A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 34 m/s2 for the time interval t2 = 49 s.
(a) Enter an expression for the rocket's speed, v1, at time t1 in terms of the variables provided.
(b) Enter an expression for the rocket's speed, v2, at the end of the second period of acceleration, in terms of the variables provided in the problem statement.
(c) Using your expressions for speeds v1 and v2, calculate the total distance traveled, in meters, by the rocket from launch until the end of the second period of acceleration.

Answer 1

Answer:

(a) $v_1= a_1t_1$

(b) $v_2 =a_1t_1+a_2t_2$

(c) 44133.5 m

Explanation:

Given:

• $u$ = initial speed of the rocket in the first stage = 0 m/s
• $v_1$ = final speed of the rocket in the first stage
• $v_2$ = final speed of the rocket in the second stage
• $t_1$ = time interval of the first stage
• $t_2$ = time interval of the second stage
• $s_1$ = distance traveled by the rocket in the first stage

• $s_2$ = distance traveled by the rocket in the second stage
• $s$ = distance traveled by the rocket in whole time interval

Part (a):

Since the rocket travels at constant acceleration.

$\therefore v_1 = u+a_1t_1\\\Rightarrow v_1 = a_1t_1$

Hence, the expression of the rocket's speed at time $t_1\ is\ v_1 = a_1t_1$.

Part (b):

In this part also, the rocket moves with a constant acceleration motion.

$\therefore v_2 = v_1+a_2t_2\\\Rightarrow v_2 = a_1t_1+a_2t_2$

Hence, the expression of the rocket's speed in the time interval $t_2$ is $v_2 = a_1t_1+a_2t_2$.

Part (c):

For the constant acceleration of rocket, let us first calculate the distance traveled by the rocket in both the time intervals.

$s_1 = u+\dfrac{1}{2}a_1t_1^2\\\Rightarrow s_1 = 0+\dfrac{1}{2}67\times(1)^2\\\Rightarrow s_1 =33.5\ m$

Similarly,

$s_2 = v_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = a_1t_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = 67\times1\times49+\dfrac{1}{2}\times 34\times(49)^2\\\Rightarrow s_2 =44100\ m\\\therefore s = s_1+s_2\\\Rightarrow s = 33.5\ m+44100\ m\\\Rightarrow s =44133.5\ m$

Hence, the rocket moves a total distance of 44133.5 m until the end of the second period of acceleration.

Answer 2

Answer:

(a) $v_{1}=0+a_{1}t_{1} = a_{1}t_{1}$.

(b) $v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}$

(c) $219807.5m$  

Explanation:

Part (a):

To find an expression for the rocket's speed $v_{1}$ at time $t_{1}$, we use the constant acceleration model, which relates these variables with the expression:$v_{final} =v_{intial}+at$. In this case, the initial velocity is null, because accelerates from rest. So, we take all values of the first interval, and we replace it to find the expression:

$v_{1}=0+a_{1}t_{1} = a_{1}t_{1}$. (expression for the first interval).

Part (b):

Then, we do the same process to find the expression of the second interval, we just replace the variables given:

$v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}$

In this case, you can notice that the initial velocity used is the one we obtain from the first interval, because the end of the first period is the beginning of the second period.

Part (c):

To calculate the total distance we have to sum the distance covered during the two intervals, that it's translated as: $d_{total} = d_{1} + d_{2}$.

Then, we use this equation to replace in each distance: $v_{final}^{2} = v_{initial}^{2} +2ad$.

Isolating d we have:

$d=\frac{v_{final}^{2}-v_{initial}^{2}}{2a}$.

Now, we apply the equation to each interval to obtain $d_{1}$ and $d_{2}$:

$d_{1}=\frac{v_{1}^{2}-0}{2a_{1}}$.

$d_{2}=\frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}$.

Before calculating the total distance, we need to know the magnitude of each speed.

$v_{1}=a_{1}t_{1}$

$v_{1}=(67)(39)=[tex]v_{2}=v_{1}+a_{2} t_{2}= 2613\frac{m}{s^{2}}+34\frac{m}{s^{2}}(49sec)=4279\frac{m}{s}$[/tex]

At last, we use all values known to calculate the total distance:

$d_{total} = d_{1} + d_{2}$

$d_{total} =\frac{v_{1}^{2}-0}{2a_{1}} + \frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}$.

$d_{total} =\frac{(2613)^{2} }{2(67)} +\frac{(4279)^{2}-(2613)^{2} }{2(34)}\\ d_{total}=50953.5+168854=219807.5m$  

Therefore, the total distance traveled until the ends of the second period is $219807.5m$. The rocket is in the Thermosphere.