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 4.3 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 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.
use your basic equations of motion
s = vt + 1/2 at^2
v = at
How far do you get? And just what is the question?
To solve this problem, we can use the equations of motion to find the speed and position of both the tortoise and the hare at different time intervals. Let's break down the information given step by step:
1. The hare accelerates uniformly at a rate of 0.8 m/s^2 for 4.3 seconds. We can use the formula for calculating speed with constant acceleration: v = u + at.
- Initial velocity (u) of the hare is 0 m/s (as it starts from rest).
- Acceleration (a) is 0.8 m/s^2.
- Time (t) is 4.3 seconds.
Plugging the values into the formula, we can find the final speed (v) of the hare after 4.3 seconds of acceleration.
2. The hare continues at a constant speed for 12.9 seconds. Since the speed remains constant, the hare's acceleration is 0 m/s^2. We can use the formula for calculating distance with constant speed and time: s = ut.
- Velocity (u) of the hare after acceleration is the final speed calculated in the previous step.
- Time (t) is 12.9 seconds.
Plugging the values into the formula will give us the distance (s) covered by the hare during this period.
3. The hare then slows down with constant acceleration and comes to a stop after traveling 62 meters. We can use the equation of motion: v^2 = u^2 + 2as.
- Initial velocity (u) of the hare is the final speed calculated in the first step.
- Final velocity (v) is 0 m/s (as the hare comes to rest).
- Distance (s) is 62 meters.
Plugging the values into the equation will help us find the acceleration (a) experienced by the hare during this period.
4. The tortoise accelerates uniformly for the entire distance and finally catches up to the hare. We need to find the time it takes for the tortoise to catch up to the hare at the exact moment the hare comes to a stop.
- The distance covered by the hare after the constant speed period can be calculated by adding the distance covered during the acceleration period and the distance covered during the deceleration period.
- Let's denote this distance as D.
- The acceleration experienced by the tortoise is not provided, so we need to calculate it using the formula: a = (v^2 - u^2) / (2s), where v and u are the initial and final velocities, and s is the distance covered during acceleration.
- Once we have the acceleration of the tortoise, we can calculate the time it takes to catch up with the hare by using the formula: t = sqrt(2s / a).
By following these steps and performing the necessary calculations, we can find the speed, distance, acceleration, and time it takes for the tortoise to catch up to the hare.