Consider a skateboarder who starts from rest at the top of a 12.5m-long ramp that is inclined at an angle of 15.0degrees to the horizontal. Assuming that the skateboarder's acceleration is g sin 15.0degrees, find his speed when he reaches the bottom of the ramp in 3.00 seconds

That would be V = a t, where a is the acceleration rate (which is given) and t = 3.00 s. Using this method, I get

V = a t = 7.6 m/s

But they also give you the length of the ramp X, and the acceleration rate. From that, one gets
V = sqrt (2 a X) = 8.0 m/s

They have given you more data than you need and the data are somewhat inconsistent.

Well, well, well! We have a gravity-loving skateboarder zooming down a ramp. That sounds like a fun ride! Let's calculate his speed, shall we?

First things first, we need to find the acceleration of our adventurous skater. We know that acceleration is equal to the gravitational acceleration (g) times the sine of the angle of the incline (15 degrees). So, his acceleration will be g * sin(15 degrees).

Now we can compute the acceleration value. Since I am a clown who avoids math, I'll leave the actual numerical value as g * sin(15 degrees).

Next, we need to find his final velocity after he grooves his way down the ramp. We can use the equation of motion:

v = u + at

Where:
v is the final velocity
u is the initial velocity (which is zero in this case since he starts from rest)
a is the acceleration
t is the time taken (3 seconds).

Substituting our values in, we get:

v = 0 + (g * sin(15 degrees)) * 3

But wait, we still need the value for g. In good ol' Earth, the acceleration due to gravity is approximately 9.8 m/s². So, let's replace g with 9.8 and do some calculations:

v = 0 + (9.8 * sin(15 degrees)) * 3

Now, grab your calculator and do the math for me, my friend. Calculate that expression, and you'll have your final answer for how fast our skater is barreling down that ramp!

Remember, physics is your friend... but laughter is the ultimate ride!

To find the speed of the skateboarder when he reaches the bottom of the ramp, we can use the following steps:

Step 1: Calculate the displacement of the skateboarder along the ramp.

We can use the equation:

displacement = initial velocity * time + 0.5 * acceleration * time^2

Here, the initial velocity is 0 (since the skateboarder starts from rest), the time is 3.00 seconds, and the acceleration is g * sin(15.0°).

So we have:

displacement = 0 * 3.00 + 0.5 * (9.8 m/s^2) * (sin 15.0°) * (3.00)^2

displacement = 0 + 0.5 * 9.8 * (0.259) * 9

displacement = 11.627 m

Step 2: Calculate the final velocity of the skateboarder when he reaches the bottom.

We can use the equation:

final velocity = initial velocity + acceleration * time

Again, the initial velocity is 0, the acceleration is g * sin(15.0°), and the time is 3.00 seconds.

So we have:

final velocity = 0 + (9.8 m/s^2) * (sin 15.0°) * 3.00

final velocity = 0 + 9.8 * (0.259) * 3

final velocity = 7.869 m/s.

Therefore, the skateboarder's speed when he reaches the bottom of the ramp after 3.00 seconds is approximately 7.869 m/s.

To find the skateboarder's speed when he reaches the bottom of the ramp, we can use the equation of motion:

v = u + at

Where:
- v represents the final velocity
- u represents the initial velocity (which is 0, as the skateboarder starts from rest)
- a represents the acceleration
- t represents the time

Here, the acceleration is given by g sin θ, where g is the acceleration due to gravity (approximately 9.8 m/s²) and θ is the angle of inclination (15.0 degrees).

So, we have:
a = g sin θ = (9.8 m/s²) * sin(15.0°)

Next, substituting the values we have into the equation, we get:

v = 0 + (9.8 m/s²) * sin(15.0°) * 3.00 s

Now, we solve the equation:

v = (9.8 m/s²) * sin(15.0°) * 3.00 s

Using a calculator, we find:

v ≈ 7.10 m/s

Therefore, the skateboarder's speed when he reaches the bottom of the ramp is approximately 7.10 m/s.