An object travels North at #9 m/s# for #4 s# and then travels South at #3 m/s# for # 1 s#. What are the object's average speed and velocity?

Question

An object travels North at #9  m/s# for #4 s# and then travels South at #3  m/s# for #  1   s#.  What are the object's average speed and velocity?

Emily Abbate · Accepted Answer

#"average speed" = 7.8 m/s# and #"average velocity" = 6.6 m/s#

While going North, the object goes #9 m/cancel(s) * 4 cancel(s) =36 m#.

While going South, the object goes #3 m/cancel(s) * 1 cancel(s) =3 m#.

For average speed , direction does not matter, just total distance and total time.

#"average speed" = "total distance"/"total time" = (36 m+3 m)/(4 s + 1 s) = 7.8 m/s#

For average velocity, direction does matter, so we need the resultant of all displacements (which are vectors) and total time.

#"average velocity" = "net displacement"/"total time" = (36 m-3 m)/(4 s + 1 s) = 6.6 m/s# to the North.

I hope this helps, Steve

Charlotte Aaron · Answer

undefined The object's average speed is $ \frac{{\text{{total distance}}}}{{\text{{total time}}}} = \frac{{(9 \, \text{m/s} \times 4 \, \text{s}) + (3 \, \text{m/s} \times 1 \, \text{s})}}{{4 \, \text{s} + 1 \, \text{s}}} = \frac{{36 \, \text{m} + 3 \, \text{m}}}{{5 \, \text{s}}} = 7.8 \, \text{m/s} $.

The object's average velocity is $ \frac{{\text{{total displacement}}}}{{\text{{total time}}}} = \frac{{(9 \, \text{m/s} \times 4 \, \text{s}) - (3 \, \text{m/s} \times 1 \, \text{s})}}{{4 \, \text{s} + 1 \, \text{s}}} = \frac{{36 \, \text{m} - 3 \, \text{m}}}{{5 \, \text{s}}} = 6.6 \, \text{m/s} $ north.