A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 0.8 m/s^2 for 4.3 seconds. It then continues at a constant speed for 12.9 seconds, before getting tired and slowing down with constant acceleration coming to rest 62.0 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. 1) How fast is the hare going 3.4 seconds after it starts?
2) How fast is the hare going 11.3 seconds after it starts?
3) How far does the hare travel before it begins to slow down?
4) What is the acceleration of the hare once it begins to slow down?
5) What is the total time the hare is moving?
6) What is the acceleration of the tortoise?

Explanation:

To solve this, we start by using one of the equations of motion. The very first one, in fact

1

V = U + at.

V = 0 + 0.8 * 3.4 = 2.72 m/s.

2.

V = 0 + 0.8 * 4.3 = 3.44 m/s.

3.

d = ½ * 0.8 * 4.3² + 3.44 * 12.9

d = 7.396 + 44.376

d = 51.77 m.

4.

d = 62 - 51.77 = 10.23 m. = Distance

traveled during deceleration.

a = (V² - Vo²) / 2d.

a = (0² - 3.44²) / 20.46

a = -11.8336 / 20.46 = -0.58 m/s²

5.

t = (V - Vo)/a =(0 - 3.44) / -0.58

t = -3.44/-.58 = 5.93 s

= Stop time.

T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total

time the hare was moving.

6.

d = Vo * t + ½ * a * t² = 62 m.

0 + 0.5 * (23.13)² * a = 61

267.5a = 61

a = 61/267.5

a = 0.23 m/s²

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 .

the ima (higher than 1m) will be greater than the ima with the height of 1m.