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 1.5 m/s^2 for 4.1 seconds. It then continues at a constant speed for 11.9 seconds, before getting tired and slowing down with constant acceleration coming to rest 98 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 1.2 seconds after it starts?
2) How fast is the hare going 12.8 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?
Thanks a lot!
1,2) v = v0 + at
3) s = v0*t + 1/2 at^2
see what you can do with those, remembering to change values at t=4.1
what do you come up with?
To answer these questions, we need to analyze the motion of the hare and the tortoise separately. Let's start with the hare:
1) To find the speed of the hare 1.2 seconds after it starts, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the hare starts from rest, so u = 0. Since the hare accelerates uniformly at a rate of 1.5 m/s^2, we have a = 1.5 m/s^2. Plugging in the values, we get v = 0 + 1.5 * 1.2 = 1.8 m/s. So, the hare is moving at a speed of 1.8 m/s.
2) To find the speed of the hare 12.8 seconds after it starts, we need to consider two different phases of its motion: acceleration and constant speed. First, let's find the speed during the acceleration phase. We can use the equation of motion: v = u + at. Again, the initial velocity u is 0 because the hare starts from rest. The acceleration a is given as 1.5 m/s^2, and the time t is 4.1 seconds. Plugging in the values, we have v = 0 + 1.5 * 4.1 = 6.15 m/s. So, during the acceleration phase, the hare reaches a speed of 6.15 m/s.
After the acceleration phase, the hare continues at a constant speed for 11.9 seconds, which means its speed does not change. Therefore, the speed of the hare at 12.8 seconds is still 6.15 m/s.
3) To find the distance traveled by the hare before it begins to slow down, we need to calculate the total distance traveled during the acceleration and constant speed phases. During acceleration, we can use the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. The initial velocity u is 0, the acceleration a is 1.5 m/s^2, and the time t is 4.1 seconds. Plugging in the values, we get s1 = 0 + (1/2) * 1.5 * (4.1)^2 = 8.0625 m.
During the constant speed phase, the hare continues for 11.9 seconds at a speed of 6.15 m/s. We can use the equation s = vt, where s is the distance, v is the speed, and t is the time. Plugging in the values, we get s2 = 6.15 * 11.9 = 73.185 m.
So, the total distance traveled by the hare before it begins to slow down is s1 + s2 = 8.0625 + 73.185 = 81.2475 m.
4) The acceleration of the hare once it begins to slow down can be found using the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. Since the hare comes to rest, the final velocity v is 0. The initial velocity u is given as 6.15 m/s (the speed during the constant speed phase). The distance s is given as 98 m. Plugging in the values, we can solve for a:
0^2 = (6.15)^2 + 2 * a * 98
Simplifying the equation: 0 = 37.8225 + 196a
Rearranging the equation: a = -37.8225/196 ≈ -0.193 m/s^2
So, the acceleration of the hare once it begins to slow down is approximately -0.193 m/s^2.
5) To find the total time the hare is moving, we need to sum the time during the acceleration phase and the time during the constant speed phase. The time during the acceleration phase is given as 4.1 seconds. The time during the constant speed phase is given as 11.9 seconds. Adding these two values together, we get 4.1 + 11.9 = 16 seconds. Therefore, the hare is moving for a total time of 16 seconds.
Now let's move on to the tortoise:
6) The acceleration of the tortoise is given as uniformly accelerating over the entire distance. Since the tortoise starts from rest and catches up with the hare just as the hare comes to a stop, the acceleration of the tortoise is equal to the deceleration (negative acceleration) of the hare. Therefore, the acceleration of the tortoise is approximately -0.193 m/s^2.