Question

A spaceship ferrying workers to moon Base i takes a straight-line pat from the earth to the moon, a distance of 384,000km, Suppose the spaceship starts from rest and accelerates at 20.0 meter per second squared (m/s2).for the first 15.0min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 meter per second squared (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:

95.78125%

18000 m/s

22233.33 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+20\times 15\times 60\\\Rightarrow v=18000\ m/s$

The velocity of the rocket at the end of the first 15 minutes is 18000 m/s which is the maximum speed of the rocket in the complete journey.

Distance traveled while speeding up

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 20\times 900^2\\\Rightarrow s=8100000\ m$

Distance traveled while slowing down

$s=18000\times 900+\dfrac{1}{2}\times -20\times 900^2\\\Rightarrow s=8100000\ m$

Distance traveled during constant speed

$384000000-(2\times 8100000)=367800000\ m$

Fraction

$\dfrac{367800000}{384000000}\times 100=95.78125\ \%$

Fraction of the total distance is traveled at constant speed is 95.78125%

Time taken at constant speed

$t=\dfrac{367800000}{18000}=20433.33\ s$

Total time taken is $900+20433.33+900=22233.33\ s$