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Anastaziya [24]
3 years ago
13

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

of the resulting velocity?
a)

14 m/s South


b)

14 m/s North


c)

10 m/s North


d)

2 m/s North
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

v=8+(-6)\\\\v=2\ m/s

It is positive. So, it is in North direction.

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<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

y(t)=y_{0}-\frac{1}{2}gt^2

Where y(t) is the vertical distance at a time t, y_{0} is the initial vertical distance: y_{0}=10m And g is the gravitational acceleration: g=9.81m/s^2 }

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The time is the same as in the last question: t=0.8s

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2

y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)

y(0.8)=10m-\frac{1}{2}(6.2784m)

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(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using y(t)=y_{0}-\frac{1}{2}gt^2

if y(t)=0

0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t

and since g=9.81m/s^2

t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s

At t=1.43s the car hits the water, so the horizontal distance at that time is:

x(t)=x_{0}+v_{0}t

x(1.43)=(2m/s)(1.43s)=2.86m

<u>the answer to (c) is 2.86s</u>

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