Question

An unsuspecting bird is coasting along in an easterly direction at 3.00 mph when a strong wind from the south imparts a constant acceleration of 0.400 m/s2 . If the wind's acceleration lasts for 2.90 s , find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally travelling in the +x direction and that there are 1609 m in 1 mi .

r=

mθ=

∘

Now, assume the same bird is moving along again at 3.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a ϕ=36.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.600 m/s2 , find the displacement vector →r and the angle of the displacement θ1 .

Enter the components of the vector, rx and ry , and the angle. Assume the time interval is still 2.90 s .

→r=rx^i+ry^j

rx=

mry=

mθ1=

∘

Answer 1

given velocity = 3 mph = 1.34 m/s

in vector form, v_o = 1.34 i + 0 j

and

acceleration is given as 0 i + 0.4 j

now, we can use kinematics to find the final position

r = v_ot + 1/2at²

r = (1.34 i + 0 j ) 2.90 + 1/2 * ( 0i + 0.4 j ) 2.90²

r = (3.886 i +0 j) + ( 0 i + 1.682 j )

r = 3.886 i + 1.682 j

so,

magnitude of position

r = sqrt ( 3.886² + 1.682²)

r = 4.234 m

and

direction can be found as

= arctan ( 1.682 / 3.886)

= 23.4 degree ( North of east)

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Now, in second case

v_o = 1.34 i + 0 j

a = 0.6 * cos 36 i + 0.6 * sin 36 j

a = 0.4854 i + 0.3526 j

so,

again , use the same equation as we did in part (1)

r = v_ot + 1/2at²

r = (1.34 i + 0 j ) 2.90 + 1/2 * (0.4854 i + 0.3526 j ) 2.90²

r = (3.886 i +0 j) + (2.0411 i + 1.482 j)

r = 5.9271 i + 1.482 j

so,

magnitude of r = 6.1 m

and

= 14.038 degree ( north of east)