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/s2 for 4.4 seconds. It then continues at a constant speed for 13.5 seconds, before getting tired and slowing down with constant acceleration coming to rest 61.0 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 2.2 seconds after it starts?

2. How fast is the hare going 15.1 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?

1. V = Vo + at.

V = 0 + 0.8m/s^2*2.2s = 1.76 m/s.

2. V = 0 + 0.8*4.4 = 3.52 m/s = Velocity
4.4 sec. after it start = the Velocity 15.1 s. after it starts.

3. d = 0.5*0.8*(4.4)^2 + 3.52*13.5 = 55.3 m.

4. d = 61 - 55.3 = 5.7 m. = Distance
traveled during deceleration.

a = (V^2-Vo^2) / 2d.
a = (0-(3.52)^2 / 11.4 = -1.09 m/s^2.

5.t = (V-Vo)/a =(0-3.52) / -1.09=3.23s.
= Stop time.

T = 4.4 + 13.5 + 3.23 = 21.1 s. = Total
time the hare was moving.

6. d = Vo*t + 0.5a*t^2 = 61 m.
0 + 0.5*(21.1)^2*a = 61
222.61a = 61
a = 0.274 m/s^2.

post it.

To solve these questions, we'll need to use physics principles related to acceleration, velocity, and motion equations. Let's go step by step to find the answers.

1. To find the hare's speed 2.2 seconds after it starts, we can use the equation:
v = u + at

Given:
u (initial velocity) = 0 (as the hare starts from rest)
a (acceleration) = 0.8 m/s²
t (time) = 2.2 seconds

Plugging in these values, we have:
v = 0 + (0.8 * 2.2)
v = 1.76 m/s

Therefore, the hare is moving at a speed of 1.76 m/s 2.2 seconds after it starts.

2. To find the hare's speed 15.1 seconds after it starts, we need to consider the two phases of its motion: the phase of acceleration and the phase of constant speed.

Phase 1: Acceleration
To find the final velocity at the end of the acceleration phase, we can use the equation:
v = u + at

Given:
u (initial velocity) = 0 (as the hare starts from rest)
a (acceleration) = 0.8 m/s²
t (time) = 4.4 seconds

Plugging in these values, we have:
v1 = 0 + (0.8 * 4.4)
v1 = 3.52 m/s

Phase 2: Constant Speed
The hare maintains this speed for 13.5 seconds. Therefore, its speed remains constant at v1 (3.52 m/s) during this time.

Therefore, the hare's speed 15.1 seconds after it starts is 3.52 m/s.

3. To find the distance the hare travels before it begins to slow down, we can use the equation:
s = ut + (1/2)at²

Given:
u (initial velocity) = 0 (as the hare starts from rest)
a (acceleration) = 0.8 m/s²
t (time) = 4.4 seconds

Plugging in these values, we have:
s = 0 + (1/2)(0.8)(4.4)²
s = 0 + (0.4)(19.36)
s = 7.744 meters

Therefore, the hare travels 7.744 meters before it begins to slow down.

4. To find the acceleration of the hare once it begins to slow down, we can use the equation:
v² = u² + 2as

Given:
u (initial velocity) = 3.52 m/s (speed at the end of acceleration phase)
v (final velocity) = 0 (as the hare comes to rest)
s (distance) = 61 meters (given in the problem)

Plugging in these values, we have:
0² = (3.52)² + 2(a)(61)
0 = 12.3904 + 122a
-12.3904 = 122a
a = -0.1012 m/s²

Therefore, the acceleration of the hare once it begins to slow down is -0.1012 m/s² (negative value indicates deceleration).

5. To find the total time the hare is moving, we need to consider the different phases of its motion.

Phase 1: Acceleration - 4.4 seconds
Phase 2: Constant Speed - 13.5 seconds
Phase 3: Deceleration until coming to rest - t seconds (to be determined)

The total time T is the sum of these phases:
T = 4.4 + 13.5 + t

Using the distance traveled during the deceleration phase, we can calculate t:
s = ut + (1/2)at²

Given:
u (initial velocity) = 3.52 m/s (speed at the end of acceleration phase)
a (acceleration) = -0.1012 m/s² (deceleration)
t (time) = to be determined
s (distance) = 61 meters (given in the problem)

Plugging in these values, we have:
61 = 3.52t + (1/2)(-0.1012)(t)²
61 = 3.52t - 0.0506t²
0.0506t² - 3.52t + 61 = 0

Solving this quadratic equation will give us the time t. Using the quadratic formula or factoring, we can find t. The positive root will be the correct time.

6. To find the acceleration of the tortoise, the problem states that it accelerates uniformly for the entire distance and catches the hare just as it comes to a stop. This indicates that the tortoise's acceleration is equal to the hare's acceleration during the first phase.

Therefore, the acceleration of the tortoise is 0.8 m/s².

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