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yulyashka [42]
2 years ago
14

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

After traveling in a straight line for 2.13 km, they stop and discover that they have been traveling 25.9 ° north of east, because their guide misread his compass.
What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?
Physics
1 answer:
aleksandrvk [35]2 years ago
5 0

Exactly, it is a vector subtraction problem.  Let t=theta,

t1=38.7, d1=4.63;  t2=25.9, d2=2.13

v1=d1 <cos(t1), sin(t1)>=<3,6134, 2.8949>

v2=d2 <cos(t2), sin(t2)>=<1.9161, 0.9304>

Final vector

v3 = v1-v2

=<v3x, v3y>

=<1.6973, 1.9645>

where

v3x=d1*cos(t1)-d2*cos(t2)

v3y=d1*sin(t1)-d2*sin(t2)

The final vector v3 has therefore a magnitude of

||v3||=sqrt(1.6973^2+1.9645^2)=2.5962, and a direction of

theta=atan(1.9645/1.6973)=49.17 degrees  north of east

Note: all three vectors are in the direction of the first quadrant.

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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

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