An object travels North at #15 m/s# for #2 s# and then travels South at #2 m/s# for #8 s#. What are the object's average speed and velocity?

Answer 1

speed = 4.6 m/s
velocity = 1.4 m/s (north) or -1,4m/s (south)

To start of, we have to find the the total distance (or disposition covered) north and south which can be derived from the speed (or velocity) formula and total time taken #speed="distance"/"time"#
therefore, #"distance"=speed*time#
When the object is traveling north: #"distance"=15*2=30 meters#
When the object is traveling south #"distance"=2*8=16 meters#
#Total time = 2+ 8=10 seconds#

Now that we are aware of the differences between speed and velocity, we must make that distinction.

Speed is #"total distance covered" / "time"#

however,

Velocity is #"total disposition"/"time"#

Disposition is a vector quantity, whereas distance is a scalar quantity, hence the two are not the same.

Since scalar quantities lack direction, calculations are performed without them.

#"distance"= 30 meters + 16 meters = 46 meters# thus #"speed"=46/10=4.6 "m/s"#

However, direction affects vector quantities, so we must verify the directions in our computations.

#"disposition"= 30 "meters (north)" -16 "meters (south)" = 14 "meters (north)" # (the minus sign is because north is opposite direction to south) thus, #"velocity"=14/10=1.4 "m/s" (north)#
when calculated to south, the velocity will be #16-30=-14 "m/s south"#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The object's average speed is ( \frac{{\text{{total distance}}}}{{\text{{total time}}}} = \frac{{(15 , \text{m/s} \times 2 , \text{s}) + (2 , \text{m/s} \times 8 , \text{s})}}{{2 , \text{s} + 8 , \text{s}}} = \frac{{30 , \text{m} + 16 , \text{m}}}{{10 , \text{s}}} = \frac{{46 , \text{m}}}{{10 , \text{s}}} = 4.6 , \text{m/s}). The object's average velocity is ( \frac{{\text{{total displacement}}}}{{\text{{total time}}}} = \frac{{0 , \text{m/s} - 15 , \text{m/s}}}{{2 , \text{s} + 8 , \text{s}}} = \frac{{-15 , \text{m/s}}}{{10 , \text{s}}} = -1.5 , \text{m/s}).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7