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 19300km/h to 1600km/h in 4.0min
Stage B: A parachute then opened to slow it down to 321km/h in 94s
Stage C: Retro rockets then fired to reduce its speed to zero over a distance of 75m
Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant.

part 1) Find the rocket's acceleration (in m/s^2) during stages A,B,and C.

Part 2) Find the total distance the rocket traveled (in km) during stage A,B,and C.

A) Divide the velocity change by the time interval cases A and B. For part C, where you are given the distance and not the time interval,

time interval = 2x/V ; a = V^2/(2x)
There will be a need to convert speeds to m/s

B) Multipy the average (mean) velocity of each stage by the corresponding time interval, and add them up.

Please show more effort if you wish further help.

Part 1:

To find the acceleration during each stage, we can use the equation of motion:

v = u + at

Where,
v = final velocity
u = initial velocity
a = acceleration
t = time

Stage A: From 19300 km/h to 1600 km/h in 4.0 min, which is equal to 4.0 * 60 = 240 seconds.

Initial velocity (u) = 19300 km/h = 19300 * 1000 / 3600 = 5361.11 m/s
Final velocity (v) = 1600 km/h = 1600 * 1000 / 3600 = 444.44 m/s
Time (t) = 240 seconds

Using the equation v = u + at, we can find the acceleration (a) during Stage A:

444.44 = 5361.11 + a * 240

Simplifying the equation:

a * 240 = 5361.11 - 444.44
a * 240 = 4916.67
a = 4916.67 / 240
a ≈ 20.49 m/s^2

Therefore, the acceleration during Stage A is approximately 20.49 m/s^2.

Stage B: From 321 km/h to 0 km/h in 94 seconds.

Initial velocity (u) = 321 km/h = 321 * 1000 / 3600 = 89.17 m/s
Final velocity (v) = 0 m/s
Time (t) = 94 seconds

Using the equation v = u + at, we can find the acceleration (a) during Stage B:

0 = 89.17 + a * 94

Simplifying the equation:

a * 94 = -89.17
a = -89.17 / 94
a ≈ -0.95 m/s^2

Therefore, the acceleration during Stage B is approximately -0.95 m/s^2.

Stage C: From a velocity of 0 km/h to 0 km/h over a distance of 75m.

Initial velocity (u) = 0 m/s
Final velocity (v) = 0 m/s
Distance (s) = 75 meters

Using the equation v^2 = u^2 + 2as, we can find the acceleration (a) during Stage C:

0 = 0 + 2 * a * 75

Simplifying the equation:

2 * a * 75 = 0
a * 150 = 0
a = 0 / 150
a = 0 m/s^2

Therefore, the acceleration during Stage C is 0 m/s^2.

Part 2:

To find the total distance traveled during Stage A, B, and C, we need to sum up the distances traveled during each stage.

Distance traveled during Stage A:
Using the equation v = u + at:

v = u + at
444.44 = 5361.11 + 20.49 * t

Solving for t:

20.49t = 444.44 - 5361.11
20.49t = -4916.67
t = -4916.67 / 20.49
t ≈ -240 seconds

Since time (t) cannot be negative, we'll take the absolute value of t: t ≈ 240 seconds

Using the equation s = ut + (1/2)at^2, we can find the distance traveled during Stage A:

s = 5361.11 * 240 + (1/2) * 20.49 * 240^2
s ≈ 1286666.4 meters
s ≈ 1286.67 km

Distance traveled during Stage B:
Using the equation s = ut + (1/2)at^2:

s = 89.17 * 94 + (1/2) * (-0.95) * 94^2
s ≈ 8367.06 meters
s ≈ 8.37 km

Distance traveled during Stage C: Given as 75 meters.

Therefore, the total distance traveled during Stage A, B, and C is approximately:

Total distance = 1286.67 km + 8.37 km + 0.075 km
Total distance ≈ 1295.12 km

So, the total distance traveled during Stage A, B, and C is approximately 1295.12 km.

Part 1) To find the rocket's acceleration during each stage, we can use the formula:

acceleration = (final velocity - initial velocity) / time taken

Stage A:
Initial velocity = 19300 km/h = 19300 * (1000/3600) m/s = 5361.11 m/s
Final velocity = 1600 km/h = 1600 * (1000/3600) m/s = 444.44 m/s
Time taken = 4.0 min = 4.0 * 60 s = 240 s

acceleration_A = (444.44 m/s - 5361.11 m/s) / 240 s = -23.28 m/s^2

Stage B:
Initial velocity = 1600 km/h = 1600 * (1000/3600) m/s = 444.44 m/s
Final velocity = 321 km/h = 321 * (1000/3600) m/s = 89.17 m/s
Time taken = 94 s

acceleration_B = (89.17 m/s - 444.44 m/s) / 94 s = -4.63 m/s^2

Stage C:
Initial velocity = 321 km/h = 321 * (1000/3600) m/s = 89.17 m/s
Final velocity = 0 m/s
Distance = 75 m

acceleration_C = (0 m/s - 89.17 m/s) / t
Distance = (1/2) * acceleration_C * t^2
75 m = (1/2) * acceleration_C * t^2

Solving for acceleration_C, we need to find the time (t) it took to reach the final velocity of 0 m/s.
Using the equation of motion: v = u + at, where v = 0 m/s, u = 89.17 m/s, we can find t.

0 = 89.17 m/s + (acceleration_C) * t
t = -89.17 m/s / acceleration_C

Substituting back into the distance equation:
75 m = (1/2) * acceleration_C * (-89.17 m/s / acceleration_C)^2
75 m = (1/2) * 89.17 m/s * (-89.17 m/s / acceleration_C)
75 m = (1/2) * (-89.17 m/s)^2 / acceleration_C
150 m = (89.17 m/s)^2 / acceleration_C
150 m * acceleration_C = (89.17 m/s)^2
acceleration_C = (89.17 m/s)^2 / 150 m
acceleration_C = 53.01 m/s^2

Part 2) To find the total distance traveled during each stage, we can use the equations of motion:

Stage A:
Distance = (1/2) * (initial velocity + final velocity) * time taken
Distance_A = (1/2) * (5361.11 m/s + 444.44 m/s) * 240 s = 933,333.6 m = 933.3336 km

Stage B:
Distance_B = (1/2) * (444.44 m/s + 89.17 m/s) * 94 s = 33,456.9 m = 33.4569 km

Stage C:
Distance_C = 75 m = 0.075 km

Therefore, the total distance traveled during each stage is:
Distance_total = Distance_A + Distance_B + Distance_C
Distance_total = 933.3336 km + 33.4569 km + 0.075 km = 966.8665 km

To find the rocket's acceleration during each stage, we need to use the kinematic equations. These equations relate the initial velocity, final velocity, acceleration, and time.

For Stage A:
Given:
Initial velocity (u) = 19,300 km/h
Final velocity (v) = 1,600 km/h
Time (t) = 4.0 minutes = 4.0 * 60 = 240 seconds

First, we convert the velocities from km/h to m/s:
Initial velocity (u) = 19,300 km/h * (1,000 m/1 km) * (1 h/3600 s) = 5,361.1 m/s
Final velocity (v) = 1,600 km/h * (1,000 m/1 km) * (1 h/3600 s) = 444.4 m/s

Using the equation v = u + at, we can solve for acceleration (a):
a = (v - u) / t
a = (444.4 m/s - 5,361.1 m/s) / 240 s
a = -4,916.7 m/s / 240 s
a ≈ -20.49 m/s^2

Therefore, the acceleration during Stage A is approximately -20.49 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.

For Stage B:
Given:
Initial velocity (u) = 1,600 km/h
Final velocity (v) = 321 km/h
Time (t) = 94 seconds

Converting velocities to m/s:
Initial velocity (u) = 444.4 m/s
Final velocity (v) = 321 km/h * (1,000 m/1 km) * (1 h/3600 s) = 89.2 m/s

Using the equation v = u + at, we can solve for acceleration (a):
a = (v - u) / t
a = (89.2 m/s - 444.4 m/s) / 94 s
a = -355.2 m/s / 94 s
a ≈ -3.78 m/s^2

Therefore, the acceleration during Stage B is approximately -3.78 m/s^2.

For Stage C:
Given:
Initial velocity (u) = 321 km/h
Final velocity (v) = 0 m/s
Distance (s) = 75 m

Converting the initial velocity to m/s:
Initial velocity (u) = 321 km/h * (1,000 m/1 km) * (1 h/3600 s) = 89.2 m/s

Using the equation v^2 = u^2 + 2as, where v = 0 m/s, we can solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (0 - (89.2 m/s)^2) / (2 * 75m)
a = -7,939.84 m^2/s^2 / 150 m
a ≈ -52.93 m/s^2

Therefore, the acceleration during Stage C is approximately -52.93 m/s^2.

Now, to find the total distance traveled during each stage:

For Stage A, we need to find the average velocity and multiply it by the time:
Average velocity = (u + v) / 2
Average velocity = (5,361.1 m/s + 444.4 m/s) / 2 = 2,902.75 m/s
Distance traveled during Stage A = Average velocity * Time
Distance traveled during Stage A = 2,902.75 m/s * 240 s = 696,660 m

Converting the distance from meters to kilometers:
Distance traveled during Stage A = 696,660 m * (1 km/1,000 m) = 696.66 km

For Stage B, the distance traveled is given as 75 m.

For Stage C, the distance traveled is also given as 75 m.

Therefore, the total distance traveled during stages A, B, and C is:
Total distance = Distance during Stage A + Distance during Stage B + Distance during Stage C
Total distance = 696.66 km + 75 m * (1 km/1,000 m) + 75 m * (1 km/1,000 m)
Total distance = 696.66 km + 0.075 km + 0.075 km
Total distance ≈ 696.81 km

Therefore, the total distance traveled during stages A, B, and C is approximately 696.81 km.