Suppose that a kid was casually sledding down a gentle slopping hill going 6 feet per second. Then, he hit a steeper part of the hill and started accelerating at a rate of 1.2 feet/second squared. At the bottom of the steep part of the hill he was going 10.56 feet/second. How long was the steep part of the this hill?
"10th grade" is not the subject
The change in speed is 9.26 ft/s.
Time spent accelerating = 9.26/1.2 = 7.72 seconds
The average speed on the steep part was 5.88 ft/s
Distance = 7.72 s * 5.88 ft/s= 45.4 feet
To find the length of the steep part of the hill, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (10.56 ft/s)
u = initial velocity (6 ft/s)
a = acceleration (1.2 ft/s^2)
s = displacement or distance (unknown)
Rearranging the equation, we can solve for s:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we have:
s = (10.56^2 - 6^2) / (2 * 1.2)
Calculating the numerator:
s = (111.5136 - 36) / (2 * 1.2)
s = 75.5136 / 2.4
s = 31.463 ft
Therefore, the steep part of the hill is approximately 31.463 feet long.