A novice captain is pointing his ferryboat directly across the river at a speed of 15.7 mi/h. If he does not pay attention to the current that is headed downriver at 5.35 mi/h, what will be his resultant speed and direction? (Consider that the current and the boat’s initial heading are perpendicular to each other.)

Anette is a civil engineer and needs to determine the length of a highway on-ramp before construc- tion begins. If the average vehicle takes 10.8 s to go from 20.0 mi/h to 60.0 mi/h, how long should she design the separate merge lane to be so a car can reach a speed of 60.0 mi/h before merging?

Vr = 5.35 + 15.7i = 16.7 mi/h[71.2o] = Resultant velocity.

Vo = 20mi/h * 1600m/mi * 1h/3600s = 8.89s.

V = 60mi/h = 26.7 m/s.

a = (V-Vo)/t = (26.7-8.89)/10.8 = 1.65 m/s^2.

d = Vo*t + 0.5a*t^2

To determine the resultant speed and direction, we can utilize vector addition.

First, let's visualize the situation. Imagine a river flowing horizontally from left to right, and the captain is traveling across the river from the bottom to the top.

The speed of the ferryboat in the absence of any current is 15.7 mi/h, and the current is flowing downstream (right) at a speed of 5.35 mi/h. Since the boat's heading and the current are perpendicular to each other, we can treat them as perpendicular vectors.

To find the resultant speed, we need to calculate the magnitude of the vector sum of the boat's velocity and the current's velocity.

Let's break down the velocities into their components:

Boat's velocity (Vb) = 15.7 mi/h (horizontal component)
Current's velocity (Vc) = 5.35 mi/h (vertical component)

Since the boat's heading and current are perpendicular, their respective components can be treated as the legs of a right triangle.

Using the Pythagorean theorem, we can calculate the magnitude of the resultant velocity (Vr):

Vr = sqrt(Vb^2 + Vc^2)

Plugging in the values, we get:

Vr = sqrt((15.7 mi/h)^2 + (5.35 mi/h)^2)
= sqrt(246.49 mi^2/h^2 + 28.6225 mi^2/h^2)
= sqrt(275.1125 mi^2/h^2)
≈ 16.58 mi/h

So, the resultant speed of the ferryboat is approximately 16.58 mi/h.

To determine the direction, we can use trigonometry. Since the boat's heading and current are perpendicular, the direction of the resultant velocity will be the angle between the resultant velocity vector and the boat's original heading.

We can find this angle (θ) using the inverse tangent (arctan) function:

θ = arctan(Vc / Vb)

Plugging in the values, we get:

θ = arctan(5.35 mi/h / 15.7 mi/h)
≈ arctan(0.3418)
≈ 19.73 degrees

Therefore, the resultant speed and direction of the ferryboat are approximately 16.58 mi/h at an angle of 19.73 degrees above the boat's initial heading.