A space shuttle is launched from rest. It takes 8.00s to reach 165km/h , and at the end of the first 1.00 min its speed is 1650km/h .

a) what is the avg. acceleration in (m/s2) during the first 8.00 seconds?

b) what is the avg. acceleration in (m/s2) between the 8.00 seconds and the first 1.00 min?

c)vAssuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the shuttle travel during the first 8.00s?

d) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the shuttle travel during the interval from 8.00s to 1.00min?

the answers are supposed to be in m/s^2

A. 5.729 m/s^2

B. 51.5625 m/s^2
C. 183.333m
D. 70,088m

a) The average acceleration during the first 8.00 seconds can be calculated using the formula:

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

Given that the initial velocity is 0 m/s and the final velocity is 165 km/h, we need to convert the final velocity to m/s:

165 km/h = 165000 m / (3600 s * 1 h) = 45.8 m/s

Plugging in the values:

average acceleration = (45.8 m/s - 0 m/s) / 8.00 s = 5.725 m/s^2

b) The average acceleration between the 8.00 seconds and the first 1.00 min can similarly be calculated using the same formula. The initial velocity is 165 km/h and the final velocity is 1650 km/h, so we need to convert them to m/s:

Initial velocity = 165 km/h = 45.8 m/s
Final velocity = 1650 km/h = 458.3 m/s

average acceleration = (458.3 m/s - 45.8 m/s) / (60 s - 8.00 s) = 11.68 m/s^2

c) To calculate the distance traveled during the first 8.00 seconds, we can use the equation:

distance = (initial velocity * time) + (0.5 * average acceleration * time^2)

Substituting the values:

distance = (0 m/s * 8.00 s) + (0.5 * 5.725 m/s^2 * (8.00 s)^2) = 183.2 m

d) To calculate the distance traveled during the interval from 8.00 seconds to 1.00 minute, we can use the same equation as above, but with different values:

distance = (165 km/h * (60 s - 8.00 s)) + (0.5 * 11.68 m/s^2 * ((60 s - 8.00 s)^2)) = 52320 m

So, the shuttle travels a distance of 183.2 m during the first 8.00 seconds, and a distance of 52320 m during the interval from 8.00 seconds to 1.00 minute.

To solve these problems, we need to use the formulas of kinematics. Let's break down each question step by step:

a) What is the average acceleration in m/s² during the first 8.00 seconds?
To find average acceleration (a_avg), we use the formula:
a_avg = Δv / Δt
where Δv is the change in velocity and Δt is the change in time.

Given:
Initial velocity (u) = 0 m/s (because the shuttle is launched from rest)
Final velocity (v) = 165 km/h = (165 * 1000) / 3600 m/s
Δt = 8.00 seconds

Now, we can calculate Δv:
Δv = v - u
Δv = (165 * 1000) / 3600 - 0 = 45.83 m/s

Substituting the values into the formula:
a_avg = Δv / Δt
a_avg = 45.83 / 8.00 = 5.73 m/s²

Therefore, the average acceleration during the first 8.00 seconds is 5.73 m/s².

b) What is the average acceleration in m/s² between the 8.00 seconds and the first 1.00 minute?
Here, we need to find the change in velocity and change in time between the 8.00 seconds and 1.00 minute.

Given:
Initial velocity (u) = 165 km/h = (165 * 1000) / 3600 m/s
Final velocity (v) = 1650 km/h = (1650 * 1000) / 3600 m/s
Δt = 1.00 minute = 60.00 seconds

Calculating Δv:
Δv = v - u
Δv = (1650 * 1000) / 3600 - (165 * 1000) / 3600 = 412.50 m/s

Substituting the values into the formula:
a_avg = Δv / Δt
a_avg = 412.50 / 60.00 = 6.88 m/s²

Therefore, the average acceleration between 8.00 seconds and the first 1.00 minute is 6.88 m/s².

c) Assuming constant acceleration, what distance does the shuttle travel during the first 8.00 seconds?
To find the distance traveled (s), we use the formula:
s = ut + (1/2)at²
where u is the initial velocity, t is the time, and a is the acceleration.

Given:
Initial velocity (u) = 0 m/s
Time (t) = 8.00 seconds
Average acceleration (a_avg) = 5.73 m/s² (from part a)

Substituting the values into the formula:
s = ut + (1/2)at²
s = 0 + (1/2) * 5.73 * (8.00)²
s = 0 + (1/2) * 5.73 * 64.00
s = 0 + 182.72
s = 182.72 m

Therefore, the shuttle travels a distance of 182.72 meters during the first 8.00 seconds.

d) Assuming constant acceleration, what distance does the shuttle travel during the interval from 8.00 seconds to 1.00 minute?
To find the distance traveled (s), we use the same formula as in part c.

Given:
Initial velocity (u) = 165 km/h = (165 * 1000) / 3600 m/s
Time (t) = 60.00 seconds
Average acceleration (a_avg) = 6.88 m/s² (from part b)

Substituting the values into the formula:
s = ut + (1/2)at²
s = (165 * 1000) / 3600 * 60.00 + (1/2) * 6.88 * (60.00)²
s = 2750 + 0.5 * 6.88 * 3600
s = 2750 + 14 784
s = 17 534 m

Therefore, the shuttle travels a distance of 17,534 meters during the interval from 8.00 seconds to 1.00 minute.

A. a = v/t 165km / 8s = 20.625 km/h/s

B. t= 60s-8s =52s, V= 1650km/h-165km/h =1485km/h
a=v/t 1485km/h / 8s= 185.625km/h/s

C. d=1/2at^2 1/2 * 20.625 km/h/s * (8s)^2 = 183.333m

D. 60s-8s=52s d=1/2at^2 1/2 * 186.625km/h/s * (52s)^2 = 70.088km