A sled starts from rest at the top of a hill and slides down with a constant acc
ID: 1953269 • Letter: A
Question
A sled starts from rest at the top of a hill and slides downwith a constant acceleration. At some later time it is 14.4 m from
the top; 2.00 s after that it is 25.6 m from the top, 2.00 slater
40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
(a) What is the magnitude of the average velocity of the sled
during each of the 2.00-s intervals after passing the 14.4-m point?
(b) What is the acceleration of the sled? (c) What is the speed
of the sled when it passes the 14.4-m point? (d) How much time
did it take to go from the top to the 14.4-m point? (e) How far did
the sled go during the first second after passing the 14.4-m point?
Explanation / Answer
a.) The average velocities are: First 2 seconds after: (25.6m - 14.4m) / 2.00s = 5.6 m/s Next 2 seconds after: (40.0m - 25.6m) / 2.00s = 7.2 m/s Next 2 seconds after: (57.6m - 40.0m) / 2.00s = 8.8 m/s b.) To solve for the acceleration, you can use the differences in average velocities you found in part a. 8.8 m/s - 7.2 m/s = 1.6 m/s Since there is constant acceleration, the acceleration is the change in velocities over the time interval, or 1.6 m/s / 2 s = 0.8 m/s^2. c.) To get the speed at the 14.4m mark, you can use the acceleration you calculated above in this equation: Vf^2 = Vi^2 + 2 * a * d where Vf is the speed we want, Vi = 0 (since the sled started from rest), a = 0.8 m/s^2, and d = 14.4 m. Vf = sqrt(23.04) m/s d.) To get the time it took to get to the 14.4 m mark, you can use the equation: Vf = Vi * t + a * t where Vf is what we calculated above, Vi = 0, and a = 0.8 m/s^2 t = sqrt(23.04) / 0.8 e.) To get the distance traveled within the first second after the 14.4 m mark, you can use the following equation: d = Vf * t + (1/2) * a * t^2 where d is the distance we want, Vf is what we calculated above, t = 1 s, and a = 0.8 m/s^2 d = sqrt(23.04) + 0.4