The submarine V.K. Konovalov launches a torpedo at the Red October , which is 67
ID: 1879897 • Letter: T
Question
The submarine V.K. Konovalov launches a torpedo at the Red October, which is 6732 meters distant. Unfortunately, the Red October cannot run away because it has had a reactor accident. After being ejected from the torpedo tube with an initial speed of 3.94 m/s, the torpedo accelerates with a constant acceleration of 0.196 m/s2 for 95.7 s, after which it continues on with uniform motion.
(a) What is the total time required for the torpedo to reach the Red October from the moment it was launched? (answer: 336 s)
(b) After the torpedo reaches its final speed, by how much does the distance between it and the Red October decrease in ten seconds? (answer: 227 m)
(c) If this question is on the exam, you will also be asked to sketch the position versus time (y versus t), velocity versus time (v versus t) and acceleration versus time (a versus t) graphs for the torpedo's motion from the moment it was launched until the moment it hits the Red October.
Please show which equations used to solve the problem.
Explanation / Answer
a)
consider the motion during acceleration
vo = initial velocity = 3.94 m/s
a = acceleration = 0.196 m/s2
t = time taken = 95.7 sec
distance travelled during acceleration is given as
x = vo t + (0.5) a t2
x = (3.94) (95.7) + (0.5) (0.196) (95.7)2
x = 1274.6 m
d = total distance of red october = 6732 m
xo = remaining distance = d - x = 6732 - 1274.6 = 5457.6 m
v = = final constant velocity = vo + at
v = (3.94) + (0.196) (95.7) = 22.7 m/s
time taken during constant velocity motion
t' = xo /v = 5457.6/22.7 = 240.42 sec
T = total time taken = t + t' = 95.7 + 240.42 = 336 s
b)
d = distance travelled
v = constant velocity gained = 22.7 m/s
t = time of travel = 10 s
distance travelled is given as
d = v t
d = 22.7 x 10
d = 227 m
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