udent/Assignment t?dep 17044817 (c) At the end of the 60-second period, how far

Question

udent/Assignment t?dep 17044817 (c) At the end of the 60-second period, how far is the conmuter train ahead of the freight train? Note: you can enter a number Ske 1.23 x 1024 on WebAssign as 1.23e4 As of September, 2017, NASA's Dawn spacecraft is nearing the end of its asteroid belt, between Mars and Jupiter, sending back plenty of excellent photos decade-long mission. You can read more about it here. It orbited two protoplanets (Vesta and Ceres) in the spacecraft to rely on ion propulsion. Ions are accelerated by high voltage to high speeds, propelling the spacecraft in the opposite direction. The n can be rather small. Let's say that the spacecraft accelerates from rest to 65 miles per hour over a time period of 7 days. Assuming the acceleration is constant, what is the magnitude of the acceleration? 480e-5m/s/s (b) Despite this small acceleration, fairly high speeds can be reached, and significant distances covered, by having the s pacecraft accelerate for a long time. How far would Dawn travel, starting from rest, with the acceleration above taking place for 880 days? Submit Answer Save Progress My Notes Ask Your T 33 points I Previous Answers ype here to search

Explanation / Answer

Given,

v = 65 mph = 29.06 m/s ; t = 7 days = 604800 sec

We know that, acceleration is, a = v/t

a = 29.06/604800 = 4.8 x 10^-5 m/s^2

Hence, a = 4.8 x 10^-5 m/s^2

b)we have, a and t = 880 days = 7.6 x 10^7 s

we know from eqn of motion

S = ut + 1/2 at^2

S = 0 + 0.5 x 4.8 x 10^-5 x (7.6 x 10^7)^2 = 1.38 x 10^11 m

Hence, S = 1.38 x 10^11 m

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.