Your English teacher can swim with a constant speed of 0.67m/s. While facing south , he attempts to swim straight across the St. Lawrence River , which moves east at 1.4 m/s
a) find his velocity as seen from the riverbank.
B) if the river was 175m wide
How long does it take him to cross the
River
Hri
teacher: v1=0.67 m/s,
river: v2 =1.4 m/s,
v=sqrt(v1²+v2²) =1.55 m/s. (to the east of south)
L =175 m,
s=?
v/v1=s/L =>
s= v•L/v1 =1.55•175/0.67 =404 m.
t=s/v =404/1.55 =260 s =4.34 min
To find the teacher's velocity as seen from the riverbank, we need to consider the vector addition of his swimming velocity and the velocity of the river.
a) The teacher is swimming south at 0.67 m/s, while the river is moving east at 1.4 m/s. Since these velocities are at right angles to each other, we can use the Pythagorean theorem to find the resultant velocity:
Velocity = √((velocity of teacher)^2 + (velocity of river)^2)
Velocity = √((0.67 m/s)^2 + (1.4 m/s)^2)
Velocity = √(0.4489 m^2/s^2 + 1.96 m^2/s^2)
Velocity ≈ √2.4089 m^2/s^2
Velocity ≈ 1.55 m/s
Therefore, the teacher's velocity as seen from the riverbank is approximately 1.55 m/s.
b) To determine how long it takes the teacher to cross the river, we can use the equation:
Time = Distance / Velocity
Since we know that the width of the river is 175 m and the velocity of the teacher as seen from the riverbank is 1.55 m/s, we can substitute these values into the equation:
Time = 175 m / 1.55 m/s
Time ≈ 112.9 seconds
Therefore, it takes approximately 112.9 seconds for the teacher to cross the river.