Velocity and acceleration of a turtle

[tiffanyg]

a turtle is moving with a constant acceleration along a straight ditch. He starts a stopwatch as the turtle passes a fence post and notes that it takes him 10s to reach a pine tree 10m farther along a ditch. as the turtle passes the pine tree, his speed is 1.2m/s.

how far was he from the fence post when he started from rest??

 

[jackanese]

8m i think

first find the velocity at teh fecne post (u = 2a/s - v) then find the acceleration, which is constant for the entire trip, then it's a matter of using the info to find the total distance traveled and minus that from 10.

 

[kerimek]

I would first like to clarify that the velocity and acceleration of a turtle are dependent on various factors such as its size, weight, and the surface it is moving on. However, assuming that this turtle is moving with a constant acceleration along a straight ditch, we can use the given information to calculate its velocity and acceleration.

From the given information, we know that the turtle took 10 seconds to cover a distance of 10m, which means its average velocity was 1m/s (10m/10s). Furthermore, we know that at the end of the 10 seconds, the turtle's speed was 1.2m/s. This means that its velocity was increasing at a constant rate of 0.2m/s every second, which is its acceleration.

Now, to answer the question of how far the turtle was from the fence post when it started from rest, we can use the equation of motion: s = ut + 0.5at^2, where s is the distance, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

Plugging in the values, we get s = 0 + 0.5(0.2)(10)^2 = 10m. This means that the turtle was 10m away from the fence post when it started from rest.

In conclusion, the turtle had an initial velocity of 0m/s, an acceleration of 0.2m/s^2, and was 10m away from the fence post when it started from rest. This information can help us better understand the turtle's movement and track its future progress.