A powerboat heads due northwest at 14 m/s relative to the water across a river that flows due north at 4.8 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?

13.15m/s, N 20.052°W

Explanation: The diagramatic representation of this question has been attached to this answer, please kindly find below because references will be made to it.

Due to the fact that the boat is moving northwest with a speed of 14m/s and the speed of the water flows at 4.8m/s due north, they can be represented using triangular vectors.

From the attachment below.

H = 14m, O= 4.8m and A=?

Using phythagoras theorem, we have that

H² =O² + A²

14² = 4.8² + A²

A² = 14² - 4.8²

A² = 144 - 23.04

A² = 120.96

A= √120.96

A= 13.15m

The direction of the vector is gotten by using the fact that

tan θ = O/A

tan θ = 4.8/14

tan θ = 0.3650

θ = tan^-1 (0.3650)

θ= 20.052°

From the attachment, the vector is lying north east west, hence direction is N 20.052° W

It’s a) absolute dating

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.