in anniversary celebration of Marilyn bell's 1954 feat - she was the first person to swim across lake Ontario - a swimmer set out from the shores of New York and maintained a velocity of 4 m/s [N]. as the swimmer approached the Ontario's shore, she encountered a cross-current of 2 m/s [E 25˚S]. calculate her velocity with the respect to the crowd from the beach (with respect to the ground).
add the two velocities:
v = <0,4> + <4cos25°,-4sin25°> = <x,y>
All angles are measured CW from +y-axis.
Vr = 4m/s[0o] + 2m/s[115o],
Vr = (4*sin0+2*sin115) + (4*cos0+2*cos115)i,
Vr = 1.81 + 3.15i = 3.63m/s[29.9o] CW. = resultant velocity.
To calculate the swimmer's velocity with respect to the ground, we need to consider both their initial velocity and the effect of the cross-current.
1. Start with the swimmer's initial velocity of 4 m/s [N]. This means the swimmer is moving directly towards the north.
2. Next, we need to break down the cross-current into its northward and eastward components. The given direction is E 25˚S, which means the cross-current has a component towards the east (25 degrees south of east). To find the eastward component of the cross-current, we can calculate:
Eastward component = Cross-current velocity * cos(angle)
Eastward component = 2 m/s * cos(25°) ≈ 1.8018 m/s
3. Now we can combine the swimmer's initial velocity and the eastward component of the cross-current using vector addition. Since the swimmer is moving north and the eastward component is towards the east, we can add the magnitudes of the velocities in each direction:
Velocity with respect to the ground = √(Northward velocity² + Eastward velocity²)
Velocity with respect to the ground = √(4 m/s)² + (1.8018 m/s)²
Velocity with respect to the ground ≈ √(16 m²/s² + 3.2452 m²/s²)
Velocity with respect to the ground ≈ √(19.2452 m²/s²)
Velocity with respect to the ground ≈ 4.3875 m/s
Therefore, the swimmer's velocity with respect to the ground (the crowd on the beach) is approximately 4.3875 m/s.