A hare is resting behind a rock, when he suddenly sees a tortoise cruising by at a constant 4.0m/s. The hare, literally at rest initially, decides he must catch up with the tortoise, to save face. So, he accelerates uniformly to 6.0m/s in 4.0 seconds. Once he reaches 6.0m/s, he continues on at constant velocity indefinitely.

a) How fast is the hare traveling 2.0 seconds after he sets off?

(a) The hare is traveling 12 m/s after seconds. Just multiply acceleration by time since there was a standing start.

If there are more parts to this question, please attempt them yourself before posting again.

To find how fast the hare is traveling 2.0 seconds after he sets off, we can break the problem into two parts: the acceleration phase and the constant velocity phase.

During the acceleration phase, the hare increases its speed from 0 m/s to 6.0 m/s uniformly in 4.0 seconds. To find the acceleration during this phase, we can use the formula:

acceleration = change in velocity / time

The change in velocity is 6.0 m/s - 0 m/s = 6.0 m/s, and the time is 4.0 seconds. Therefore, the acceleration is:

acceleration = 6.0 m/s / 4.0 s = 1.5 m/s^2

Next, we can use the formula for uniformly accelerated motion to find the velocity after 2.0 seconds during the acceleration phase:

final velocity = initial velocity + acceleration * time

The initial velocity is 0 m/s (since the hare starts from rest), the acceleration is 1.5 m/s^2, and the time is 2.0 seconds. Plugging in these values, we have:

final velocity = 0 m/s + 1.5 m/s^2 * 2.0 s = 3.0 m/s

So, 2.0 seconds after the hare sets off, during the acceleration phase, the hare is traveling at a speed of 3.0 m/s.

During the constant velocity phase, the hare continues to travel at 6.0 m/s indefinitely. Therefore, if 2.0 seconds have elapsed, the hare is still traveling at 6.0 m/s.

So, the final answer is that 2.0 seconds after the hare sets off, it is traveling at a speed of 3.0 m/s.