An airplane covers a straight-line distance of 8.72km in 35.9 s, during which time it has a constant forward acceleration of 3.03 m/s^2. a) what is the speed of the airplane, in m/s, at the beginning of the 35.9 s? b) what is the speed of the airplane, in m/s, at the end of the 35.9 s?

Explanation:

It is given that,

Distance covered by the airplane,

Time taken, t = 35.9 s

Acceleration of the airplane,

(a) Let u is the initial speed of the airplane at the beginning of the 35.9 seconds. It can be calculated using the second equation of motion as :

u = 188.50 m/s

(b) Let v is the speed of the airplane at the end of the 35.9 seconds. It can be calculated using the first equation of motion as :

v = 297.27 m/s

Hence, this is the required solution.

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.  

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over