If you divide the total distance travelled by a car trip by the time are you calculating the average speed or the magnitude of the average velocity? Under what circumstances are these two quantities the same? Illustrate with the help of an example.

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Solution

If you divide the total distance travelled on a car trip by the time for the trip, then you are calculating average speed.

Average velocity is given by displacement divided by time taken. So magnitude of average velocity will be equal to average speed only when displacement of the object is equal to the distance travelled by the object. And it happens when the object moves in a straight line in one direction only.

For example, Let's say point B is 10m north to Point A in a straight line. If you are travelling along the straight path from point A to point B, then your displacement is same as your distance travelled, that is 10m. So in this case your average speed and magnitude of average velocity will be same

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Similar questions

Q. Statement 1: On a curved path average speed of a particle can never be equal to average velocity.
Statement 2: Average speed is total distance travelled divided by total time. Whereas average velocity is, final velocity plus initial velocity divided by two.

Explain clearly, with examples, the distinction between:
a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].

A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 k m h^{- 1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5 k m h^{- 1}$ . What is the
a) magnitude of average velocity, and
b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min,
(iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

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Q. Which of the following statements are incorrect?
(i) Average velocity is path length divided by time interval.
(ii) In general, average speed is greater than the magnitude of the average velocity.
(iii) A particle moving in a given direction with a non-zero velocity can have zero speed.
(iv) The magnitude of average velocity is the average speed.

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