# Average Speed and Free-Falling Objects

• Soaring Crane

## Homework Statement

A car is making a trip of 40 mi. It travels half the distance at an average speed of 20 mph. In order for it to have an average speed of 40 mph for the whole trip, the car would need to:

a. travel at an average speed of 40 mph for the trip’s remainder.
b. travel at an average speed of 60 mph for the trip’s remainder.
c. cover the remainder of the distance in 15 minutes.
d. It is not possible for the entire trip to have an average speed of 40 mph..

## Homework Equations

average speed = distance / delta time

## The Attempt at a Solution

Based on the mathematical definition of average speed, the time taken to travel the first 20 mi is:

Time = distance / average speed = 20 mi / 20 mph = 1 hr

The time required to travel the desired distance, 40 mi, with an average speed of 40 mph is:

Time = distance / average speed = 40 mi / 40 mph = 1 hr

For the first 20 mi, 1 hr has already elapsed; therefore, the total time to travel 40 mi will be longer than 1 hr. Is the correct choice d.?

## Homework Statement

An object at rest is dropped from a height of 10 m. After 1 s, what is the object’s speed? (g = 10 m/s^2)

v_f = v_o +a*t

## The Attempt at a Solution

v_f = v_o +a*t

v_o = 0 m/s
t = 1 s
a = -10 m/s^2

v_f = (-10 m/s^2)*(1 s) = |-10 m/s| = 10 m/s ?

Thank you.

The car problem is correct.

For the dropped object, you might or might not be correct. What is the object's speed after it has fallen 10 m?

I don't quite understand your posed question. Wouldn't the object be at rest once it lands 10 m below the building?

Or do you want me to find v_f with the equation (v_f)^2 = (v_o)^2 + 2*a*(delta x) in which v_f = sqrt[2*(-10 m/s^2)*-10 m]?

I don't quite understand your posed question. Wouldn't the object be at rest once it lands 10 m below the building?

Yes, exactly. So the answer is either 0, or the 10 m/s you calculated. It depends on whether the object has fallen 10m in 1 s.