time, both engines tum on. The one on the left gives the of v, -3555 m/s and v,
ID: 1879527 • Letter: T
Question
time, both engines tum on. The one on the left gives the of v, -3555 m/s and v, 4150 m/s. what was the magnitude and the direction of the spacecrat's intial velooty before the engines were turned on? Express the magnitue as m/s and the airection efrom the +x ais magnitude Readng Oider stered by magnete hii. Co Ider one wch electron that is emitted with an·tal veocity of 205 x 107 m s in the horrontal area on s.00 x 1015 mis? The phosphorescent screen is a horzontal distance of 5.6 m away from the point where the electron is emitted. magnetic forces de let the electron with a vertically upward acceleration of How much time does the electron take to travel from re emanon D to re screen?Explanation / Answer
vfx = 3555m/s, vfy =4150m/s
ax = 5.10m/s2, ay = 7.30m/s2
t=635s
Use kinematic equations as below,
vfx = vix +ax*t
vix = vfx - ax*t = 3555- 5.10*635 = 316.5 m/s
vfy = viy +ay*t
viy = vfy - ay*t = = 4150- 7.30*635 = -485.5 m/s
magnitude = |vi|= sqrt(viy2 + viy2 ) = sqrt(485.5^2+911.5^2) = 1032.7 m/s
direction = = tan^-1(vfy/vfy) = tan^-1(485.5/316.5) = 56.9o clockwise from +x axis
= (360-56.9) = 303.1o counterclockwise from +x axis
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