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An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +35 ft/s2. After some time t1...Asked by Candace
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +42 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. Assume the total distance traveled by the sled is 15800 ft and the total time is 90 s.
a) find times t1 & t2 (s)
b)find velocity (ft/s)
At the 15800 ft mark, the sled begins to accelerate at -24 ft/s2.
(c) What is the final position of the sled when it comes to rest?
ft
(d) What is the duration of the entire trip?
a) find times t1 & t2 (s)
b)find velocity (ft/s)
At the 15800 ft mark, the sled begins to accelerate at -24 ft/s2.
(c) What is the final position of the sled when it comes to rest?
ft
(d) What is the duration of the entire trip?
Answers
Answered by
drwls
t1 + t2 = 90
15,800 ft = acceleration distance + coasting distance
(1/2)*42*t1^2 + (42*t1)*t2 = 15800
Solve those two equations in the two unknowns, t1 and t2. The substitution method should work.
(b) V = 42 * t1 (after acceleration)
Use the value of t1 derived in part (a) to compute the initial V when deceleration begins.
(c) Let t' = deceleration time
24 t' = V
Solve for t'
Additional distance travelled while decelerating = (V/2)*t'
Add that to 15,800 for final position.
15,800 ft = acceleration distance + coasting distance
(1/2)*42*t1^2 + (42*t1)*t2 = 15800
Solve those two equations in the two unknowns, t1 and t2. The substitution method should work.
(b) V = 42 * t1 (after acceleration)
Use the value of t1 derived in part (a) to compute the initial V when deceleration begins.
(c) Let t' = deceleration time
24 t' = V
Solve for t'
Additional distance travelled while decelerating = (V/2)*t'
Add that to 15,800 for final position.
Answered by
King Jew
jewish!!!
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