Velocity Physics Example 5

Follow the full solution, then compare it with the other examples linked below.

Example 5

medium
A car drives 120 km120 \text{ km} north in 1.5 hours1.5 \text{ hours}, then 80 km80 \text{ km} south in 1 hour1 \text{ hour}. Find: (a) average speed, (b) average velocity.

Solution

  1. 1
    (a) Average speed = total distance / total time = 120+801.5+1=2002.5=80 km/h\frac{120 + 80}{1.5 + 1} = \frac{200}{2.5} = 80 \text{ km/h}.
  2. 2
    (b) Net displacement = 120−80=40 km north120 - 80 = 40 \text{ km north}. Average velocity = 402.5=16 km/h north\frac{40}{2.5} = 16 \text{ km/h north}.

Answer

(a)  80 km/h;(b)  16 km/h north(a)\; 80 \text{ km/h}; \quad (b)\; 16 \text{ km/h north}
Average speed uses total distance (always positive); average velocity uses net displacement (direction matters). When an object reverses direction, velocity is much less than speed.

About Velocity

The rate of change of position with respect to time, including both magnitude and direction.

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More Velocity Examples

Example 1 easy

A car travels [formula] east in [formula]. What is its average velocity?

Example 2 medium

A cyclist rides [formula] north in [formula], then [formula] south in [formula]. What is the average

Example 3 medium

A particle's position is given by [formula]. What is the instantaneous velocity at [formula]?

Example 4 medium

A car accelerates from [formula] to [formula] in [formula]. Find the average velocity.

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