a sled starts from rest and slides down a 38.0 degree frictionless hill for 4.24 seconds. How fast is it going at the end

acceleration is ... g * sin(38.0º)

velocity = acceleration * time

To find the final velocity of the sled at the end of its slide down the hill, we can use the equation for constant acceleration:

v = u + at

Where:
v = final velocity
u = initial velocity (0 m/s, as the sled starts from rest)
a = acceleration
t = time (4.24 seconds)

First, we need to find the acceleration of the sled.

The acceleration of an object sliding down an incline can be calculated using the formula:

a = g * sin(θ)

Where:
a = acceleration
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the incline (38 degrees)

Substituting the given values, we have:

a = 9.8 m/s^2 * sin(38)

Now, let's calculate the acceleration:

a = 9.8 m/s^2 * sin(38) ≈ 5.971 m/s^2

Now that we have the acceleration, we can calculate the final velocity:

v = 0 + (5.971 m/s^2) * (4.24 s)

v ≈ 25.315 m/s

Therefore, the sled is going approximately 25.315 m/s at the end of its slide down the hill.

To find the speed of the sled at the end of the hill, we need to use the principles of physics, specifically the equations of motion.

First, we can use the components of the sled's initial velocity to find its initial speed along the hill. Since the sled starts from rest, its initial velocity along the hill is zero.

Next, we can use the acceleration due to gravity and the distance traveled on the hill to find the final speed. Since the hill is frictionless, the only force acting on the sled is the force due to gravity. This force causes the sled to accelerate along the hill as it slides down.

The equation that relates the final velocity with the initial velocity, acceleration, and time is:

v = u + at

where:
v = final velocity (unknown)
u = initial velocity (zero)
a = acceleration (due to gravity, g = 9.8 m/s^2)
t = time (4.24 seconds)

Plugging in the values:

v = 0 + 9.8 * 4.24

v ≈ 41.45 m/s

Therefore, the sled is moving at approximately 41.45 m/s at the end of the hill.