Cars A and B are traveling around the circular race track. At the instant shown,
ID: 1594996 • Letter: C
Question
Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 75 ft/s and is increasing its speed at the rate of 15 ft / s2, whereas B has a speed of 101 ft/s and is decreasing its speed at 25 ft / s2.
Part A
Determine the relative velocity of car A with respect to car B at this instant.
Express your answer to three significant figures and include the appropriate units.
Part B
Determine the angle v of vA/B (counterclockwise from the negaitive x axis).
Express your answer to three significant figures and include the appropriate units.
Part C
Determine the relative acceleration of car A with respect to car B at this instant.
Express your answer to three significant figures and include the appropriate units.
Part D
Determine the angle a of aA/B (counterclockwise from the positive x axis).
Express your answer to three significant figures and include the appropriate units.
vA/B = A 300 ft 60 B 250 ftExplanation / Answer
given facts:
speed of car A=75 ft/sec
acceleration=+15 ft/s^2
speed of car B=101 ft/s
acceleration=-25 ft/sec^2
let i and j are unit vectors along +ve x and +ve y axis
then velocity of car A=Va=(-75 i) ft/sec
acceleration of car=Aa=(-15 i) ft/sec^2
velocity of car B=Vb=101*(-cos(60) i + sin(60) j)
=-50.5 i + 87.469 j
acceleration of car B=Ab=-25*(-cos(60) i+sin(60)j )=12.5 i -21.651 j
centripetal acceleration of car A:
magnitude=speed^2/radius=75^2/300=18.75 ft/s^2
it is radially inwards
so in vector notation, total acceleration=aA=-15 i - 18.75 j ft/s^2
centripetal acceleration of car B:
magnitude=speed^2/radius=101^2/250=40.804 ft/s^2
direction is radially inwards
hence in vector notation, centripetal acceleration=40.804*(-sin(60) i- cos(60) j)
=-35.337 i - 20.402 j
total acceleration of B=aB=12.5 i -21.651 j -35.337 i -20.402 j=-22.837 i -42.053 j
part A:
relative velocity of A w.r.t B=Va-Vb=(-75 i)-(-50.5 i + 87.469 j)
=-24.5 i -87.469 j
magnitude=sqrt(24.5^2+87.469^2)
=90.835 ft/s
part B:
angle with -ve x axis=arctan(87.469/24.5)
=74.352 degrees
part C:
relative acceleration of A w.r.t. B=aA-aB=-15i-18.75 j +22.837 i +42.053 j
=7.837 i + 23.303 j
magnitude=sqrt(7.837^2+23.303^2)=24.586 ft/s^2
part D:
angle with +ve x axis=arctan(23.303/7.837)=71.412 degrees