Question

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?

Answer 1

Answer:

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