A sled starts from rest at the top of a hill and slides down with aconstant acce

Question

A sled starts from rest at the top of a hill and slides down with aconstant acceleration. At some later time it is 24.4 m from the top; 1.50seconds after that it is 33.4 m from thetop, 1.50 seconds later 43.9 m from the top and 1.50seconds later it is 55.7 m from the top. (a) What is the magnitude of the averagevelocity of the sled during each of the 1.50 s intervals after passing the 24.4 m point?
first interval
1 m/s
second interval
2 m/s
third interval
3 m/s

(b) What is the acceleration of the sled?
4 m/s2 (down the hill)

(c) What is the speed of the sled when it passes the 24.4 m point?
5 m/s

(d) How much time did it take to go from the top to the24.4 m point?
6 s

(e) How far did the sled go during the first second after passingthe 24.4 m point?
7 m (a) What is the magnitude of the averagevelocity of the sled during each of the 1.50 s intervals after passing the 24.4 m point?
first interval
1 m/s
second interval
2 m/s
third interval
3 m/s

(b) What is the acceleration of the sled?
4 m/s2 (down the hill)

(c) What is the speed of the sled when it passes the 24.4 m point?
5 m/s

(d) How much time did it take to go from the top to the24.4 m point?
6 s

(e) How far did the sled go during the first second after passingthe 24.4 m point?
7 m

Explanation / Answer

2 ( 43.9m - 33.4m)/(1.5s) = 7m/s
3 ( 55.7m - 43.9m)/(1.5s) = 7.86m/s ˜8m/s
4 Acceleration a = (7m/s - 6m/s)/(1s)       = 1m/s2
5 v = [(2)(1m/s2)(24.4m)]    = 6.98m/s
6 t = v/a (since initial velocity is zero)    =(6.98m/s)/(1m/s2 )    = 6.98 m/s
7 y = (6.98m/s)(1s) +(1/2)(1m/s2)(1s)2    = 7.48s

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