Question

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Suppose it accelerates at an acceleration 20.2 m/s2 for the first time interval 15.8 min of the trip, then travels at constant speed until the last time interval 15.8 min , when it accelerates at − 20.2 m/s2 , just coming to rest as it reaches the moon.

(a) What is the maximum speed attained?
(b)What fraction of the total distance is traveled at constant speed?
(c)What total time is required for the trip?"

Answer 1

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$

Answer 2

Acceleration of spaceship is change of its speed in the trip of Earth to the moon.

• (a) The maximum speed attained is 19149.6 m/s.
• (b) Fraction of the total distance is traveled at constant speed is 95 percent.
• (c)Total time required for the trip is 350 minutes.

What is the acceleration of spaceship?

Acceleration of a spaceship is the rate of change of velocity of the it per unit time.

Given information-

Distance of Earth to the moon is 384,000 km.

The acceleration of the spaceship for first time interval of 15.8 min or 948 seconds is 20.2 meter per squared second.

The acceleration of the spaceship for last time interval of 15.8 or 948 seconds min is -20.2 meter per squared second.

• (a) The maximum speed attained-

As the initial velocity of the spaceship is zero. Thus the final velocity can be given as,

$v_f=0+20.2\times948\\v_f=19149.6\rm m/s$

Hence, the maximum speed attained by the spaceship is 19149.6 m/s.

• (b) Fraction of the total distance is traveled at constant speed-

Total distance traveled by the spaceship in first time interval with 20.2 meter per squared second is,

$s_{first}=\dfrac{19149.6\times948}{2}\\s_{first}=9076.91$

Total distance traveled by the spaceship in last time interval with -20.2 meter per squared second is,

$s_{last}=\dfrac{19149.6\times948}{2}\\s_{last}=9076.91$

Total distance traveled with constant speed is,

$s=s_{total}-s_{first}-s_{last}\\s=384000-9076.91-9076.91\\s=365846.2\rm m$

Percentage of this distance to the total distance is,

$\Delta s= \dfrac{365846.2}{380000}\\\Delta s\cong0.95$

Thus the fraction of the total distance is traveled at constant speed is 95 percent.

• (c)Total time is required for the trip-

Total time required to travel with constant speed is,

$t=\dfrac{365856.2\times100}{19149.6}\\t=19100\rm s\\t=318.3\rm min$

Thus the total time of the trip is,

$t=15.8+318.3+15.8\\t\cong350\rm min$

Thus, total time required for the trip is 350 minutes.

Hence,

• (a) The maximum speed attained is 19149.6 m/s.

• (b) Fraction of the total distance is traveled at constant speed is 95 percent.

• (c)Total time required for the trip is 350 minutes.

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