1. You are in a kayak which can move at a velocity of 2.6 m/s when in still wate

Question

1. You are in a kayak which can move at a velocity of 2.6 m/s when in still water. You are attempting to cross a river which is 110 across and which is flowing to the East at a speed of 0.4 m/s. You head directly North across the river.

a. How long does it take you to get to the other side?

b. How far downriver are you when you reach the other side of the river?

c. If you now head back upriver, how long does it take you to reach a point on the river opposite from where you started?

2. You are playing raquetball at the rec center. When you are 10 m away from the wall you hit a ball 30 m/s towards the wall at an angle of 30 degrees. Ignore any affects due to air resistance.

a. How long is the ball in the air before it hits the wall?

b. How high above the launch point does the ball hit the wall?

Explanation / Answer

When the student swims upstream, the speed of the current subtracts from the student's normal swimming speed. So the student's speed (relative to the riverbank) is:

speed_upstream = (swim_speed) - (river_speed) =

Use the usual formula (time = distance/speed) to figure the time it takes to go upstream:

time_upstream = distance / speed_upstream

110/2.6 =42.30769230769231 sec

b) in this time, you would have drifted down the river by some distance, given by:
(velocity along the flow of river) * (time)

= 0.4* 42.3= 16.92307692307692 sec

When the student swims downstream, the speed of the current ADDS to the student's swimming speed:

speed_downstream = (swim_speed) + (river_speed) =

So the time for the downstream trip is:

time_downstream = distance / speed_downstream

Finally, just add time_upstream + time_downstream to get the total time.

> How much longer or shorter will the trip be if the river is standing still?

Hah! A "river standing still" is an odd concept. But anyway: no current, so the student's speed is just the same as their swimming speed:

time_there = distance / swimming_speed
time_back = distanct / swimming_speed

total_time = time_there + time_back

c)

C)
In order to reach a point exactly opposite to the starting point, velocity along flow of the river must be zero
let us say velocity of boat along flow = Vx and that along width = Vy
Since these two are mutually perpendicular directions,
Vx2 + Vy2 must be equal to V2 where V is the velocity of the boat in still water.

Vx2 + Vy2 = 2.22

in order for velocity along flow to be zero, Vx must equal 0.4

Therefore,

0.42 + Vy2 = 2.62

therefore, Vy2 = 6.6

Vy = 2.56 m/s

Therefore time taken to reach opposite point

= width/Vy = 110/2.56 = 42.8174419289s

2)

x direction
x = v0x t
10 = 30*cos(30)*t
t=1.35375 s

b) y = y0 + v0y t + 1/2 a t^2

y = 0 + 30*sin(30)*1.375 -0.5*9.81*1.375^2=11.35m

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