(I'm only posting the questions I couldn't figure out, don't worry.)

An object released from rest at time t=0 slides down a frictionless incline a distance of 1 meter during the first second. The distance traveled by the object during the time interval from t=1 second to t=2 seconds is...

Its average speed is distance over time, or 1 m/s. Since its initial speed is zero, its final speed vf1 is 2m/s to get that average.

Now, the acceleration:
a=changevelocity/time. Compute acceleration.

Now, assume that acceleration is constant.

distance1to2= distanceattime2-distanceattime1
= 2m/s*1sec+1/2 a*2^2 - (1/2 a*1^2)

To determine the distance traveled by the object during the time interval from t=1 second to t=2 seconds, we need to consider the motion of the object on the incline.

Given that the incline is frictionless, we can assume that the only force acting on the object is due to gravity, which is directed down the incline. This force can be resolved into two components: one parallel to the incline and the other perpendicular to the incline.

Since the incline is assumed to be frictionless, the perpendicular component of the gravitational force does not contribute to the motion of the object. Therefore, only the parallel component of the gravitational force is responsible for the acceleration of the object down the incline.

The acceleration of the object can be determined using the formula a = gsinΞΈ, where g is the acceleration due to gravity (which is approximately 9.8 m/s^2) and ΞΈ is the angle of the incline with respect to the horizontal.

In this problem, we are not given the angle ΞΈ, so we cannot calculate the exact acceleration. However, we can assume that the angle is small enough that the acceleration is approximately equal to g.

Therefore, the acceleration of the object down the incline during the entire period from t=0 to t=2 seconds is approximately 9.8 m/s^2.

To calculate the distance traveled by the object during this time interval, we can use the formula for the distance traveled by an object under constant acceleration:

d = ut + (1/2)at^2,

where d is the distance traveled, u is the initial velocity (which is 0 in this case), t is the time interval, and a is the acceleration.

Plugging in the values, we get:

d = 0 + (1/2)(9.8)(2-1)^2
= 0 + (1/2)(9.8)(1)^2
= (1/2)(9.8)(1)
= 4.9 meters.

Therefore, the distance traveled by the object during the time interval from t=1 second to t=2 seconds is 4.9 meters.

To find the distance traveled by the object during the time interval from t=1 second to t=2 seconds, we can use the equation for distance traveled by an object under constant acceleration:

𝑑 = 𝑒𝑑 + 1/2π‘Žπ‘‘Β²

where:
- 𝑑 is the distance traveled
- 𝑒 is the initial velocity (which is zero since the object is released from rest)
- 𝑑 is the time interval
- π‘Ž is the acceleration

In this case, we are given that the object slides down a frictionless incline, which means that the only force acting on it is gravity. The acceleration of the object down the incline is equal to the acceleration due to gravity, which is approximately 9.8 m/sΒ².

Let's plug the values into the formula:

𝑑 = 0*(1) + 1/2 * 9.8 * (1)Β²

Simplifying the equation:

𝑑 = 0 + 1/2 * 9.8 * 1

𝑑 = 0 + 4.9

𝑑 = 4.9 meters

Therefore, the object travels a distance of 4.9 meters during the time interval from t=1 second to t=2 seconds.