l\'\' 1-13. A space ship zips past Mars\' orbit accelerates in a at a speed of 3
ID: 1624634 • Letter: L
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l'' 1-13. A space ship zips past Mars' orbit accelerates in a at a speed of 3 x 10 m/s and constantly straight line until it reaches the orbital path 1011 m away, when its speed is 4 x 101 m/s. of Jupiter, 5.56 x Determine A) the space ship's acceleration and (Answer in days.) B) time required from Mars" to Jupiter's orbits. P1-14. (0) Car P travels due East a constant speed of so m/s. At 9:oo along a straight highway at a.m., passes Exit 16. traveling due West P passes Exit 17. precisely the same moment. car o point Knowing At at a constant g6 later, car and car the same the exits how minutes past 9:00 a.m. the cars pass each are exactly 7 km apart, determine many other. P1-15. Seeing a traffic jam ahead, car P slams on the brakes and decelerates from so m/s at a constant rate, reaching a complete stop in 8 seconds. Determine the distance the car traveled during the braking maneuver. P 1-16. (O) Meanwhile, car Q continues at 26 m/s West and, at the next overpass, zips past a the stationary police officer, who immediately begins accelerating at 3 m/s2. Eventually, the police officer catches up to car Q, who blissfully maintains the previous speed. When police officer and care are side-by-side, the officer motions for car Q to pull over. A) Determine the time that elapsed since the police officer started up B) Determine the distance car Q traveled before being caught. e was finally nabbed C) Determine the speed the police officer was going when carExplanation / Answer
C) Speed of police = vf = vi+at = Solution:
1-13)
Acceleration = vf2- vi2 / 2d using th ekinematics formula
= (4 x10^4)^2 - (3x10^3)^2 / 2(5.56 x10^11) = 0.00143 m/s^2
Time taken = t = vf - vi / a = (40000 - 3000) / (0.00143) = 2.586 x 10^7 seconds
= (2.586 x 10^7) /(3600*24) = 300 days
1-14) Velocity of P = Vp = 30 m/s
Velocity of Q = Vq = -26 m/s
distance between the exit points = X = 7km = 7000 m
Let the two cars pass each other after a time t at some distance d from P .
distance traveled by P = d= vPt =30 t
distance traveled by Q =-(7000 - d )= Vq t = -26 t => d = 7000 - 26 t on rearranging this
equate the two equations for d
30 t = 7000 - 26 t
=> (30+26)t = 7000
=> t = 7000 /56 = 125 s = 2.1minutes
They move past each other after 2.1 min after 9AM.
1-15) distance = d = 30 t = 30 * 125 = 62.5 m = 0.0625 km
1-16) Vq = -26 m/s
Acceleration of police car = a = 3 m/s^2 . it starts from rest .
distance moved by the police car = d = 1/2 *3 t^2
distance moved by Q = (7-.0625) *10^-3- d = -26 t
=> 6937.5 +26 t =d
=> 1.5 t^2 = 6937.5 + 26 t
=> 1.5 t^2 - 26 t - 6937.5 =0
=> t = 77.2 seconds = 1.29 minutes
distance traveled by q= 6937.5 - 1.5 (77.2)^2 = -2000 m = - 2km (towards west, so negative)
c) vf = vi + at = 0 + (3)(77.2) = 231.6 m/s = 0.231 km/s
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