Question

A dragonfly flies from point A to point B along the path shown in the figure in 2.30 s.

Find the and components of its position vector at point .

What are the magnitude of its position vector at ?

What are the direction of its position vector at ?

Find the and components of the dragonfly's average velocity between and .

What are the magnitude of its average velocity between these two points?

What are the direction of its average velocity between these two points?

Answer 1

From the figurte x-component of di splacement A is 2.0 nm y-component of displacement A is 1.0 m The position of point A is (2.0 m, 1.0 m Magnitude of position of vector Ais A-20m+(1.0 m) A 2.23 m Direction of vector Ais 1.00 m 2.00 m 9-tan e 26.6 Displacement along x-direction is, x10.0 m- 2.0 m 8.0 m Average velocity along x-direction is, x 8.0 m *t 2.10 s Displacement al ong y-direction is Average velocity along y-direction is 50 m 2.38 m/s t 2.10 s Average velocity is (3.81 m/s, 2.38 m/s) Magnitude of average velocity is, Vavg-(381 m/s)+(2.38 m/s) Vavg 4.5 m/s Direction of average vel ocity is, g-tan-1/ 2.38 m/s 3.81 m/s 932°

Answer 2

a)

Position Vector of A :

x = 2 ; y = 1

b)

A = 2 i + 1 j

Magnitude of A : |A| = sqrt (2² + 1²) = sqrt (5) = 2.236

Direction of A : theta = tan ^-1 (y/x) = tan ^-1 (1/2) = 26.57^o from +x-axis.

c)

Average velocity between A and B :

Average Velocity = displacement / time

Displacement = 8 i + 5 j

Average Velocity = (8/2.3) i + (5/2.3) j = 3.478 i + 2.174 j

The x and y components of the dragonfly

Answer 3

Answer 4

a) x = +2 m

y = +1 m

b) magnitude = sqrt(x^2 + y^2) = 2.24m

direction = tan-1(y/x) = 26.56 degrees

c) Vx = 10 -2 / 2.30 = 3.48 m/s

Vy = 6 -1 / 2.30 = 2.17 m/s

d) magnitud= sqrt(Vx^2 + vy^2) = 4.10 m/s

angle= tan-1(Vy/vx) = 31.95 degrees