Questions 9 and 10 please. d dropped. Both stones vertically downward 1.o 44m ab
ID: 1539015 • Letter: Q
Question
Questions 9 and 10 please.
Explanation / Answer
Q9.
let north be along +ve y axis and east be along +ve x axis.
let i and j are unit vectors along +ve x and +vey axis.
total displacement
=600 i + 400 j +300*(sin(40) i+ cos(40) j)+700*(-cos(60) i -sin(60) j)
=(600+300*sin(40)-700*cos(60)) i + (400+300*cos(40)-700*sin(60)) j
=442.84 i + 23.596 j
part a:
average velocity=magnitude of total dispalcement/total time
=sqrt(442.84^2+23.596^2) m/(2*3600 seconds)
=0.061593 m/s
part b:
average speed=total distance/total time
=(600+400+300+700)/(2*3600) m/s
=0.27778 m/s
Q10.
let north be along +ve y axis and east be along +ve x axis.
let i and j are unit vectors along +ve x and +vey axis.
let her speed be v km/hr and angle with +ve x axis be theta.
her velocity w.r.t wind=v*(cos(theta) i + sin(theta) j)
velocity of wind w.r.t. ground=(80 km/hr) j
so velocity of the plane =velocity of plane w.r.t. wind + velocity of win w.r.t. ground
=v*cos(theta) i + (v*sin(theta)+80) j
as the direction of total velocity is 45 degree north of east
velocity along +ve x and +ve y axis should be equal.
v*cos(theta)=v*sin(theta)+80...(1)
total velocity can be written as
v*cos(theta)*(i+j)
speed=sqrt(2)*v*cos(theta)
given that speed=300 km/45 minutes
=400 km/hr
hence sqrt(2)*v*cos(theta)=400
==>v*cos(theta)=282.84 km/hr...(2)
using equation 2 in equation 1:
v*sin(theta)+80=282.84
==>v*sin(theta)=202.84....(3)
divding equation 2 by equation 3:
cot(theta)=282.84/202.84
==>theta=35.646 degrees
using the value of theta in equation 2, we get v=348 km/hr
so her speed should be 348 km/hr and direction 35.646 degree north of east.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.