a wheelbarrow travels 78.5 feet…

Question

A wheelbarrow travels 78.5 feet in 25 revolutions. What diameter is the wheel to the nearest quarter of an inch?

NOTE: The formula for this problem is S=r(theta)

S= distance, r= radius, theta= the angle in radians.

Answer 1

25 revolutions equals 50π radians

r = d / 2

78.5 = (d / 2)(50π)

Can you solve for d and answer?

Answer 2

distance divided by circumference) x pi x diameter

=78.5 feet/ 25) x diameter x pi

= 3.14 x pid

d = 3.14/3.141593 =about .9993949

= about 1 foot diameter