Question

A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 85 m/s^2 for t1 = 46 s. The first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 31 m/s^2 for the time interval t2= 41 s. Part (a) Enter an expression for the rocket's speed, v1, at time t1 in terms of the variables provided.
Part (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.
Part (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) $V_{1}$= 3910 $\frac{m}{s}$

(b) $V_{2}$= 5181 $\frac{m}{s}$

(c) $h_{t}$=  2762955 m

Explanation:

Using equations of uniformly accelerated motion:

Equations:

1. $V_{f} = V_{i} + a*t$
2. $y_{f}=y_{i} +v_{i} + \frac{1}{2} *a*t^{2}$
3. $v_{f} ^{2} =v_{i} ^{2} + 2 *a * Δy$

• $v_{i1} = 0, t_{1} = 46 s, a_{1} = 85\frac{m}{s^{2} }, a_{2} = 31\frac{m}{s^{2} }, t_{1} = 41 s$

(a) Using equation 1:

$V_{f1} = V_{i1} + a_{1}*t_{1}$ ⇒ $v_{i1}=0$ Because it upward from rest

$V_{f} = a_{1}*t_{1}$

$v_{f} = 85 \frac{m}{s^{2} } * 46 s$

$v_{f}= 3910 \frac{m}{s}$

(b) Using equation 1:

$V_{f2} = V_{i2} + a_{2}*t_{2}$ ⇒ $v_{i2}=3910$ Because is the  is the initial speed in the movement before changing acceleration

$v_{f2} = 3910 \frac{m}{s} + 31 \frac{m}{s^{2} } * 41 s$

$v_{f2} = 5181 \frac{m}{s}$

(c) Using equation 2:

$y_{f1}=y_{i1} +v_{i1} + \frac{1}{2} *a_{1}*t_{1}^{2}$  ⇒ $y_{i1}=0$ and $v_{i1}=0$ Because it upward from rest

$y_{f1}= \frac{1}{2} *a_{1}*t_{1}^{2}$

$y_{f1} = \frac{1}{2} * 85\frac{m}{s^{2} } * 46s^{2}$

$y_{f1} = 89930 m$

$y_{f2}=y_{i2} +v_{i2} + \frac{1}{2} *a_{2}*t_{2}^{2}$⇒ $y_{i2}=y_{f1}$

$y_{f2}= 89930 m + 3910 \frac{m}{s} + \frac{1}{2} * 31 \frac{m}{s^{2} } *41^{2}$

$y_{f2} = 119895,5 m$