<span>40.7 miles.
For this problem, we want to know the length of the chord created by the line and the circle. So let's first create the equations needed.
The slope intercept equation for a line is:
y = ax + b
the value for a will be the the difference in y divided by the difference in x. We're going from y=61 to y=0 for a chance of -61 and from x=0 to x=62 for a change of 62. So the value of a is -61/62, giving us the formula
y = -(61/62)x + b
Substituting x = 0, we can calculate b
61 = -(61/62)0 + b
61 = b
So the equation for the line is:
y = -(61/62)x + 61
Now for the equation for the circle. Since the circle is centered at the origin, the equation is:
x^2 + y^2 = 48^2
And we have a rather ugly quadratic equation which we can solve using the quadratic formula, giving the solutions x = 16.00512574 and x = 44.98681081
Now we need to calculate the y values for those 2 x values.
y = -(61/62)x + 61
y = -(61/62)16.00512574 + 61
y = 45.25302145
y = -(61/62)x + 61
y = -(61/62)44.98681081 + 61
y = 16.73878292
So the 2 endpoints are
(16.00512574, 45.25302145) and (44.98681081, 16.73878292)
The distance between those points can be calculated using the Pythagorean theorem.
sqrt((16.00512574 - 44.98681081)^2 + (45.25302145 - 16.73878292)^2)
= sqrt(-28.98168506^2 + 28.51423853^2)
= sqrt(839.938069 + 813.0617988)
= sqrt(1652.999868)
= 40.65710107
And finally, we have the solution of 40.7 miles.</span>
Explanation: The distance traveled is how much the ball has rolled in total, this means the lenght of the path that it has followed from begining to end.
Since it first travels 13.2 meters and then 9.5 meters, if we sum this quantities:
13.2 + 9.5 = 22.7 meters
So 22.7 is the distance that the ball has traveled.
Velocity is defined as the rate of change of displacement.
velocity is a vector quantity, that means it requires both magnitude and direction to completely explain the velocity.
For example, the velocity is 5 ms due east, it means an object is moving with speed 5 ms in the direction of east. We can say that the object covers the displacement of 5 m in one second due east.