g. /mod/ibis/view.php?id=437 3909 Gradeb o 9/20/2017 10:00 PM 0 76.9/1009/13/201

Question

g. /mod/ibis/view.php?id=437 3909 Gradeb o 9/20/2017 10:00 PM 0 76.9/1009/13/2017 11:37 AM Print R Cakuator ba Periodic Table Question 1 of 13 An unsuspecting bird is coasting along in an eastertly direction at 3.00 mph when a strong wind from the Map south imparts a constant acceleration of 0.200 m/s2. If the wind's acceleration lasts for 2.90 s, find the magnitude r and direction (measured counterclockwise from the easterly dredion) of the birds displacement over this time interval. (HINT: assume the bird is originally travelling in the *x direction and there are 1609 m in 1 mile.) Number Number e- Now, assume the same bird is moving along again at 3,.00 mph in an easterty direction but this time the acceleration given by the wind is at a 49.0 angle to the original direction of motion. If the magnitude of the acceleration is 0.300 m's?, find the displacement vector , and the angle of the displacement, 9 Enter the components of the vector and angle below. (Assume the time interval is still 2.90 s.) Number Number Number O Previos Check Answer 0 Next Exit 20

Explanation / Answer

Given,

v = 3 mph = 1.34 m/s due East ; a = 0.2 m/s6@ ; t = 2.9 s

In component form

vi = 1.34 i + 0 j

a = 0i + 0.2 j

We know from equation of motion

r = ut + 1/2 a t^2

r = 1.34 i x 2.9 + 0.5 x 0.2 j x 2.9^2

r = 3.89 i + 0.841 j

So the magnitude will be:

r = sqrt (3.89^2 + 0.841^2) = 3.98 m

theta = tan^-1(0.841/3.89) = 12.2 deg

Hence, r = 3.98 m and theta = 12.2 deg

Now, vi = 1.34 i

a = 0.2 cos49i + 0.2 x sin49 j = 0.131 i + 0.151 j

Again using the same eqn of motion:

r = 1.34 x 2.9 i + 0.5 x (0.131 i + 0.151 j)2.9^2

r = 3.89 i + 0.551 i + 0.635 j

r = 4.44 i + 0.635 j

theta = tan^-1(.635/4.44) = 8.14 deg

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