A spaceship ferrying workers to moon Base i takes a straight-line pat from the e

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???

Explanation / Answer

Given Distance between earth and moon, d = 384,000 km =3.84*10^8 m Initial velocity of the spaceship, vi = 0 m/s Acceleration of the spaceship, a = 20 m/s^2 for a time interval, t1 = 15 min = 900 s Decelaration of the spaceship ,a = - 20 m/s^2 in t2 = 15 min = 900s a) The maximum speed attained is ,     v = vi + at1     v = 0 +(20 m/s^2) (15 min) (60 s)     v = 18000 m/s is the maximum speed attained b) First, we will calculate the distance travelled during the acceleration is,     s1 = vi t1 + 1/2 at1^2          = 0 + 1/2 ( 20 m/s^2 ) (900 s)^2     s1 = 8100000 m =8.1 *10^6 m The distance travelled during the deceleration part is , s3 = v t3 + 1/2 a t3^2        = (18000 m/s) (900 s) + 1/2 (-20 m/s^2) (900 s)^2 s3 = 8100000 m = 8.1 *10^6 m Distance travelled during constant speed is , s2 = d - (s1 + s3)     = 3.84*10^8 m - (8.1 *10^6 m +8.1 *10^6 m)    s2 = 3.678 *10^8 m Fraction of total distance travelled is, (3.678*10^8 m) / ( 3.84 *10^8 m) *100 = 95.7% c) The first time interval t1 = 900 s       The last time interval , t3 = 900 s The time interval during the constant speed phase is ,     t2 = s2 / v = (3.678 *10^8 m) / (18000 m/s)     t2 = 20433.33 s Toital time taken for the trip is ,    t = t1 + t2 + t3 = 900 s +20433.33 s +900 s    t = 22233.33 s or 370.55 min

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