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Solved: A sled starts from rest at the top of a hill and
Chapter 2, Problem 66P(choose chapter or problem)
Problem 66P
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time the sled is 14.4 m from the top, 2.00 s after that it is 25.6 m from the top, 2.00 s later 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-rn 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?
Questions & Answers
QUESTION:
Problem 66P
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time the sled is 14.4 m from the top, 2.00 s after that it is 25.6 m from the top, 2.00 s later 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-rn 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?
ANSWER:
Solution to 66P
Step 1
After 14.4m the distance travelled in 2s=25.6-14.4=11.2m
Final velocity at 14.4m is the initial velocity for the next 2s. It is taken as u.The acceleration is constant and represented as ‘a’
Step2
For first 2s
11.2=2u+0.5xax(2)2
11.2=2u+2a………………………………………….(1)
For next 2s
40-14.4=4u+0.5x a x(4)2
25.6=4u+8a ………………………………………….(2)
Solving the system of equations we get,
a=0.8m/s2
u=4.8m/s
Step 3
The time taken to reach 14.4m