THIS IS MY LAST QUESTION, SO PLEASE ATTEMPT ALL QUESTIONS OR NONE AT ALL. Thanks
ID: 3280410 • Letter: T
Question
THIS IS MY LAST QUESTION, SO PLEASE ATTEMPT ALL QUESTIONS OR NONE AT ALL. Thanks.This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page find all choices before answering 004 (part 2 of 2) 10.0 points Find the position at t=58. 001 (part 1 of 2) 10.0 points 2.2-s =-8 A particle moves according to the equation z = (10 m/s) t2 where z is in meters and t is in seconds. Find the average velocity for the time i terval from 2.49 s to 4.2 s. 4. z,as =-10.5 Answer in units of m/s 002 (part 2 of 2) 10.0 points Find the average velocity for the time interval from 2.49 s to 2.59 s. Answer in units of m/s. 005 (part 1 of 2) 10.0 points Throw a ball upward from pont O with an initial speed of 62 m/s. 003 (part 1 of 2) 10.0 points An acceleration (in m/s2) has the time de- pendence shown on the graph below. The particle starts from rest (m = 0 m/s) at the origin (20 := 0 m). h . h" - Acceleration vs Time 62 m/s What is the maximu eration of gravity is -9.8 n Answer in units of m. ?The accel- 006 (part 2 of 2) 10.0 points If the speed of the ball as it passes point B is (62 m/s) = 31 m/s, what is the height hB Find the velocity at t-58. of B above O? Answer in units of m. 1. 4 m/s 007 (part 1 of 2) 10.0 points A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.8 s later with an initial speed of 74.48 m/s. They hit the ground at the same time. How long does it take the first stone to hit the ground? The acceleration of gravity is 9.8 m/s Answer in units of s.
Explanation / Answer
001
given, x = (10 m/s^2)t^2
x is meters and t in seconds
now, average velocity is given by, displacement /time
so between t1 = 2.49 s to t = 4.2 s
displacement = 10*(t2^2 - t1^2) = 114.399
time = 4.2 - 2.49 = 1.71 s
so average velocity = displacement/time = 66.9 m/s
002
x = (10 m/s/s)t^2
t1 = 2.49 s
t2 = 2.59 s
x = 10(t2^2 - t1^2) = 5.08 m
dt = t2 - t1 = 0.1 s
v = x/dt = 50.8 m/s
003
given acceleration time graph of a particle
at t = 0, vo = 0, xo = 0 m
for till t = 2s
a = -2 m/s/s
so, using v = vo + at
v(2s) = -2*2 = -4 m/s
after this time
a = 0 m/s/s
hence
v(5s) = v(2s) = -4 m/s
so option 1. -4m/s
004
for t = 2 s, a = -2 m/s/s
hence
let displacement be s
s = vo*t + 0.5at^2
s(2s) = 0.5*(-2)*4
s(2s) = -4 m
after this time
s(5s) = s(2s) + v(2s)*(5 - 2) = -4 -4*3
s(5s) = -16
hence option 3, s = -16 m
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