A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was traveling 54 m/s. Calculate (a) the acceleration, (b) the time required to travel the 150 m mentioned, (c) the time required to attain the speed of 27 m/s, and (d) the distance moved from rest to the time the train had a speed of 27 m/s.

Explanation:

A)

vf² = vi² + 2(a)(d)

(54 m/s)² = (27 m/s)² + 2(a)(120 m)

a = 9.1125 m/s²

a = 9.1 m/s²

B)

d = (vi)(t) + (1/2)(a)(t²)

150 m = 0 + (1/2)(9.1125)(t²)

t = 5.7 s

C)

vf = vi + (a)(t)

27 m/s = 0 + (9.1125)(t)

t = 3.0 s

D)

d = (vi)(t) + (1/2)(a)(t²)

d = 0 + (1/2)(9.1125)(2.96²)

d = 40 m



depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .