A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s2 until its velocity reaches 500 m/s. What distance did the plane cover while it was accelerating?

The distance the plane covered while it was accelerating is 80,633.3 m

Explanation:

Given;

initial velocity of the plane, u = 90 m/s

acceleration of the plane, a = 1.5 m/s²

final velocity of the plane, v = 500 m/s

The distance covered by the plane is given as;

v² = u² + 2ad

where;

d is the distance covered by the plane;

500² = 90² + 2(1.5)d

500²  - 90² = 3d

241900 = 3d

d = 241900 / 3

d = 80,633.3 m

Therefore, the distance the plane covered while it was accelerating is 80,633.3 m

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2