This question tests your understanding of average velocity. You have different velocities throughout your motion: you changed direction and speed several times, including periods of no motion. However, average velocity cares only about your total displacement and how long it took you to get from your initial position to your final position.

Therefore, it is unnecessary to calculate your velocity at each time interval. You know your initial position is at your house and, after the entire motion, your final position is at the bus stop which is 20m to the right of your house. From this we get that the total displacement is 20m (note that it is positive because you moved to the right).

Since you moved back and forth and were motionless for some time intervals, it took you a total of 100s to finally get to the bus stop for good and catch the bus. You need to add all the times for each section of your motion; remember there are 60s in a minute and you need to be consistent with units.

Therefore, your average velocity is:

We can determine the velocity by taking the second integral of acceleration for the time interval of 0s to 3s.

Solve for the first integral.

Solve for the second integral, using the time interval.

We can find the car's velocity by taking the integral of the acceleration function during the given time interval.

Solve the integral for the time interval of 0s to 4s. This will give us the final velocity.

Calculate the maximum acceration in meters per second.

Solve for the time.

Acceleration is equal to the derivative of the velocity function.

Since the velocity is constant, the derivative will be equal to zero.

The acceleration is equal to zero.

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

The function for acceleration is the derivative of the function for velocity.

First off we have to convert  to meters per second. 

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

Next, we use the distance equation to find the distance in the first 5 seconds:

 because the initial speed is .

If we plug in , , we get: 

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

 in this case is not equal to , it is equal to and 

Plugging in the equation, we get 

Adding  and  we get the total distance to be 

Relevant equations:

Choice I is true because  is proportional to the range , so increasing  increases  if  is constant. This relationship is given by the equation:

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon  would be less. Decreasing  increases the time the ball is in the air, thereby increasing  if  is constant. This relationship is also shown in the equation:

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation:

We know that the y-velocity is 0 to start with, acceleration is  and .

Substituting into the equation, we get:

Next we have to substitute the time into the equation for the x component

We know that  and , so we can conclude that: