a 55kg skateboarder on a 3kg board starts coasting on level ground at 8 m/sec. Let k=3.2 kg/sec. About how far will the skater coast before reaching a complete stop?
What does k represent? Air friction resistance that is proportional to velocity?
Write and solve the differential equation of motion, to get equations for X(t) and V(t). Find the t when V = 0. Use that t for the location.
The assumption that resistance is proportional to velocity may not lead to an accurate result. As velocity slows down, the frictional resistance goes to zero with that assumption.
To find out how far the skateboarder will coast before coming to a complete stop, we can use the principles of physics and the equations of motion. Specifically, we can use the equation for deceleration.
The equation for deceleration is:
a = -k * v,
where:
a is the acceleration (which is negative because it is deceleration),
k is the constant of proportionality, and
v is the velocity (speed) of the skateboarder.
In this case, k = 3.2 kg/sec and v = 8 m/sec.
First, let's find the acceleration using the equation:
a = -k * v
= -3.2 kg/sec * 8 m/sec
= -25.6 kg*m/sec^2.
Next, we can use another equation of motion to find the distance traveled (d) when decelerating from an initial velocity (u) to a final velocity (v) with a constant acceleration (a):
v^2 = u^2 + 2ad
Rearranging the equation, we have:
d = (v^2 - u^2) / (2a).
In this case, the initial velocity (u) is 8 m/sec, the final velocity (v) is 0 m/sec (since the skateboarder comes to a complete stop), and the acceleration (a) is -25.6 kg*m/sec^2.
Plugging in the values, we get:
d = (0^2 - 8^2) / (2 * -25.6)
= (-64) / (-51.2)
= 1.25 m.
Therefore, the skateboarder will coast for approximately 1.25 meters before coming to a complete stop.