Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

Answer 1
Answer:

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).


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Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s

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Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

Answers

Answer:

4.18

Explanation:

Givens  

The car's initial velocity  v_(i)= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.  

Knowns

We know that the maximum static friction force is given by:

f_(s_max) =μ_s*n                         (1)

Where μ_s is the coefficient of static friction and n is the normal force.  

Calculations  

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx=v_(i) +(1)/(2) at^2

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:  

f_(y)=n-mg=0\n n=mg\nf_(x)=F-f_(s,max) =0\n f_(s,max)=F=ma\n

Substituting (3) into (1), we get:

f_(s,max)= μ_s*m*g

Equating this equation with (4), we get:

ma=  μ_s*m*g

 μ_s=a/g

      =4.18

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Answers

Answer:

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