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 1.2 m/s2 for 4.3 seconds. It then continues at a constant speed for 11.7 seconds, before getting tired and slowing down with constant acceleration coming to rest 78.0 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.
What is the acceleration of the hare once it begins to slow down?
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To find the acceleration of the hare when it begins to slow down, we can use the formula for distance traveled, given initial velocity, acceleration, and time:
distance = initial velocity * time + (1/2) * acceleration * time^2
For the first part of the hare's motion, we are not given the initial velocity, but since it starts from rest, we know that the initial velocity is 0. We are given the acceleration (1.2 m/s^2) and the time (4.3 seconds). Let's calculate the distance covered during this time:
distance_1 = 0 * 4.3 + (1/2) * 1.2 * (4.3)^2
distance_1 = 0 + (1/2) * 1.2 * 18.49
distance_1 = 11.09 meters
Now, we know that the hare continues at a constant speed for 11.7 seconds. Since it is moving at a constant speed, its acceleration during this time is 0. Therefore, we can ignore this part when calculating the hare's final deceleration.
Next, we know that the hare comes to rest 78.0 meters from where it started. This means that the combined distances covered during acceleration and deceleration should add up to 78 meters.
Let's denote the acceleration during deceleration as "a" and the time during deceleration as "t". We need to find the values of "a" and "t" that satisfy the following equation:
distance_1 + distance_deceleration = 78
We already know the value of distance_1 (11.09 meters). So, let's calculate distance_deceleration using the formula for distance traveled during uniform acceleration:
distance_deceleration = (1/2) * a * t^2
Substituting the known values into the equation:
11.09 + (1/2) * a * t^2 = 78
Now, we need one more equation relating the time during deceleration (t) and the acceleration during deceleration (a). Since the tortoise catches the hare just as it comes to a stop, their final positions are the same. This implies that the distance covered by the hare during deceleration is the same as the distance covered by the tortoise.
The tortoise accelerates uniformly for the entire distance, so we can use the formula for distance traveled during uniform acceleration to calculate the distance covered by the tortoise:
distance_tortoise = (1/2) * acceleration * t^2
Substituting the known values:
distance_tortoise = (1/2) * a * t^2
Since the hare and tortoise cover the same distance during deceleration, we can equate the two distance equations:
(1/2) * a * t^2 = 11.09 + (1/2) * a * t^2
Simplifying the equation, we find:
11.09 = (1/2) * a * t^2
Now we have a system of two equations:
11.09 + (1/2) * a * t^2 = 78
11.09 = (1/2) * a * t^2
Solving this system of equations will give us the values of "a" and "t".