If the object is moving to the left, then its displacement must be negative since its final position will be further to the left in than its initial position. So in an imaginary number line, after some time passes, the object's final position is less than the object's initial position. This can be shown mathematically:

 , since

Since the object's displacement is negative, then so is its velocity. This can easily be seen from the equation for average velocity:

 since . At the end of its trajectory, the object stops, so overall the object's velocity is less than or equal to .

The object eventually stops. This means that the magnitude of the object's velocity is decreasing. In other words, the absolute value of the object's velocity is becoming smaller and smaller until it stops. So, for example, if at some moment the object's velocity is , after some time its velocity will be . Remember that the negative sign indicates only the direction of the vector (the object is moving to the left, hence its velocity is negative). But in absolute terms, the object is moving slower and slower (to the left) until it stops. Finally, we can see that the acceleration must be positive in order to slow down an object moving to the left. Going of our example, let's say the object is moving at and after some time it slows down to  (again the negative sign just indicates that the object moves to the left). So, if look at the average acceleration we have:

Thus the acceleration must be positive.

A trick that I often use is if something is speeding up, the acceleration and velocity point in the same direction. If something is slowing down. the acceleration and velocity point opposite directions. Since the elevator is speeding up, the acceleration must point in the same direction as it's velocity: downwards.

The origin is placed at the location the ball is thrown and caught so y=0 when it is thrown and caught. Because the ball only moves in one dimension (up or down), the velocity must drop to zero when it transitions from a upward positive velocity to a downward negative velocity at the apex. The acceleration on the surface of earth is always g for projectiles when neglecting air resistance or other forces.

The slope of a distance vs. time plot represents the object's velocity. Therefore, in both the first and second graphs, the velocity is constant and zero, respectively. An object with constant (or no) velocity is experiencing, by definition, no net force. Oppositely, in the third graph the slope is increasing, meaning the object is accelerating, meaning it must be subject to some force.

The net force experienced by m1 can be characterized as:

 where T is the tension force along the string and mg is gravity.

because  is also equivalent to  , we can express this equation as

 .

The net force experienced by m2 can similarly be expressed as:

 and  so 

From this point the problem is a system of equations.

If we solve for T in the first equation, we end up with .

Then we can plug into the second equation like so:

then:

finally we can isolate a:

when we plug in the masses given by the problem, we get:

Let's interpret each graph.  

(A) is not correct because it shows Beverly riding her bike, being in the store for a period of time, and riding back at the same speed she was going when she rode to the store. Then she goes backward in time (not possible) and ends up at the same time and location she started at.  

(B) She rides her bike to the store, stays there for a length of time and rides back.  At  (arbitrary place) she has a flat time and walks home. Notice that the slopes of the position vs. time sections before and after the store are similar. The only difference is that one is positive and one is negative, indicating a positive and negative velocity respectively. Remember that a change in position over a change in time is velocity. Now she walks home which means she is traveling slower than on her bike. This means that the slope of  vs.  is smaller. It will take her a longer time to walk back so this portion has a smaller slope. She ends up at the same  position she started from but at a later time.  

(C) This plot is correct if we were looking at her velocity vs. time. She starts out with a higher speed (positive) and then it's close to zero at the store. She then has a negative velocity that has the same magnitude as her trip to the store until she gets a flat and then her velocity is smaller but still negative as she walks home.

(D) This graph would be correct if she did not have a flat and rode home the entire way.

The sum of two vectors written in component form will result in a new vector whose components are the sum of the components of the added vectors, written as:

The magnitude of a vector is given as the square root of the sum of the squares. Written completely as:

For our example, the solution is written in full as:

By definition, the dot product of two vectors can be related to their magnitudes and the angle between them as follows:

Given the angle between the two vectors is , we can calculate the dot product to be written explicitly as:

Note that the unts are  since the dot product involves multiplying two meters together.

Displacement is not the same as distance. Over the course of the problem, the ball starts and ends in the same place. While the ball covered a distance of 14 meters during the problem, it's displacement is 0 meters.

If we say going to the right is positive and to the left is negative we have:

We calculate displacement this way because it is a vector, meaning it has both magnitude (how far it moved) and direction (which direction did it move). Distance refers only to magnitude.

Velocity and acceleration are both vectors which have a magnitude and a direction. We use the sign of a value to notate direction and in this situation we will say north is positive and south is negative. We use the equation 

Where  is the final velocity,  is the initial velocity , the acceleration is , and the time of the acceleration is .

Plug in known values and solve.

The sign of our answer is positive, meaning the car is still traveling north after the acceleration.