In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:
Stage A: Friction with the atmosphere reduced the speed from 19300 to 1600 in 4.0 .
Stage B: A parachute then opened to slow it down to 321 in 94 .
Stage C: Retro rockets then fired to reduce its speed to zero over a distance of 75 .
Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant.
A-Find the rockets acceleration (in m/s^2) for part A,B,and C
stage A: acceleartion=changevelocity/time
Stage B: ditto
Stage C: Vf^2=Vi^2+2*acceleration*distance solve for acceleration
To find the rocket's acceleration for each stage A, B, and C, we'll use the equations of motion.
For Stage A:
Given initial velocity (u) = 19,300 m/s
Final velocity (v) = 1,600 m/s
Time taken (t) = 4.0 s
Using the equation v = u + at, we can rearrange it to solve for acceleration (a):
a = (v - u) / t
Substituting the values:
a = (1,600 - 19,300) / 4.0
a = -17,700 / 4.0
a = -4,425 m/s²
Therefore, the acceleration for Stage A is -4,425 m/s².
For Stage B:
Given initial velocity (u) = 1,600 m/s
Final velocity (v) = 321 m/s
Time taken (t) = 94 s
Using the same equation, v = u + at, and rearranging it for acceleration (a):
a = (v - u) / t
Substituting the values:
a = (321 - 1,600) / 94
a = -1,279 / 94
a ≈ -13.61 m/s²
Therefore, the acceleration for Stage B is approximately -13.61 m/s².
For Stage C:
Given initial velocity (u) = 321 m/s
Final velocity (v) = 0 m/s
Distance traveled (s) = 75 m
Using the equation v² = u² + 2as, we rearrange it to solve for acceleration (a):
a = (v² - u²) / (2s)
Substituting the values:
a = (0 - 321²) / (2 * 75)
a = -102,921 / 150
a ≈ -685.47 m/s²
Therefore, the acceleration for Stage C is approximately -685.47 m/s².
To find the acceleration during each stage, we need to use the equations of motion. The equations we will use are:
1. v = u + at (Equation 1)
2. s = ut + (1/2)at^2 (Equation 2)
3. v^2 = u^2 + 2as (Equation 3)
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
- s is the distance travelled
Let's calculate the acceleration for each stage:
Stage A:
Initial velocity, u = 19300 m/s
Final velocity, v = 1600 m/s
Time taken, t = 4.0 s
Using Equation 1, we can rearrange it to find the acceleration, a:
a = (v - u) / t
a = (1600 - 19300) / 4.0
a = -17700 / 4.0
a = -4425 m/s^2 (negative because the velocity is decreasing)
So, the acceleration during Stage A is -4425 m/s^2.
Stage B:
Initial velocity, u = 1600 m/s
Final velocity, v = 321 m/s
Time taken, t = 94 s
Using Equation 1, we can rearrange it to find the acceleration, a:
a = (v - u) / t
a = (321 - 1600) / 94
a = -1279 / 94
a ≈ -13.59 m/s^2 (negative because the velocity is decreasing)
So, the acceleration during Stage B is approximately -13.59 m/s^2.
Stage C:
Initial velocity, u = 321 m/s
Final velocity, v = 0 m/s
Distance, s = 75 m
Using Equation 3, we can rearrange it to find the acceleration, a:
a = (v^2 - u^2) / (2s)
a = (0 - 321^2) / (2 * 75)
a = (-103041 - 2) / 150
a ≈ -687.61 m/s^2 (negative because the velocity is decreasing)
So, the acceleration during Stage C is approximately -687.61 m/s^2.
Therefore, the accelerations for each stage are as follows:
Stage A: -4425 m/s^2
Stage B: -13.59 m/s^2
Stage C: -687.61 m/s^2