Choice 3 is correct. Recall that physics problems tell you where to begin and end. The only thing that matters in figuring displacement is the beginning point and end point. The man effectively walked along the hypotenuse of a right triangle whose sides measured 2 km each.  Since a² + b² = c², then 4 + 4 = c², and the correct response is the square root of 8 with the direction of the displacement added for clarity.

Choice 4 is correct. In physics, it is often helpful to draw a diagram of the problem.  In this case, the respondent would draw a vector 10 hours x 10 knots from south to north (usually going towards the top of a sheet of paper) and a perpendicular vector 50 nautical miles long projecting from the arrowhead of the first vector to the left, or west. The hypotenuse of the right triangle so created is the actual path of a boat experiencing these forces. Without resorting to calculating the square root of the sum of (10,000 plus 2,500 nautical miles²), it is obvious that choice 4 is the only reasonable one.

To find total displacement, find the vector sum of the northern and eastern travel using the Pythagorean theorem.

Note that displacement differs from total distance traveled (which would be 650km).  

Since it takes the body twice the time to cover the same distance, the magnitude of v₂ is half of v₁. Also note that the direction of the body has reversed (velocity is a vector quantity), and so v₂ must be negative, with regard to v₁.

v = d/t

v₁ = d/2

v₂ = -d/4

v₂ = (-1/2)v₁

It is stated that between seconds 3 and 6 the object's acceleration increases. On a velocity vs. time graph, this would translate into a nonlinear curve.

Between seconds 1 and 3 the acceleration is constant, and would result in a linear velocity-time relationship. When the object travels at a constant velocity, the acceleration, which correlates to the slope, is zero.

Acceleration is equal to the acceleration due to gravity (about 10 m/s²) and is constant while the projectile is in the air. The force on the object is given by the product of its mass and gravitational acceleration, and is also constant.

Since the particle is decelerating the acceleration will be in the opposite direction of the movement. The particle is moving to the right, therefore its acceleration must be to the left. 

This question requires us to find the end position relative to the starting position. As a result, 150m is not the correct answer; this would be the distance, not the displacement.

The first portion involves writing down that the person moves north 100m. At this point in time, his distance on the y-axis is +100m. Next, we have to see how the person's x-axis and y-axis distances change after turning left 120^o, and walking 50m.

For the x-axis, we use the equation based on a right triangle. By turning 120^o, the man is walking at an angle of 30^o to the x-axis. Since this is the only movement in the x-axis, the total distance on the x-axis is –43.3m (movement to the left will be negative).

For the y-axis, we use the equation Since this 25m is heading downward, we subtract it from the initial upward movement of 100m. As a result, the total movement in the y-axis is 75m.

Now that we have the component vectors, we can use the pythagorean theorem to determine the total displacement of the man.

A very quick and easy way to determine the total displacement of an object is to graph the velocity of the object over time, and calculate the area under the line. The velocity vs. time graph would show a triangle with a height of  and a base of 15 seconds. The area of a triangle is determined by the equation .

As a result, we conclude that the sprinter has a total displacement of 75m.

At maximum height, the projectile is still moving in the x-direction. The horizontal component of velocity remains constant during projectile motion, therefore  .

Gravity is always acting on any body in projectile motion, therefore  . Acceleration will be constant at .