which of these cannot possibly represent one-dimensional motion of a particle

Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:

(a) The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.

(b) The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.

(c) The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.

(d) The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h -

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h ^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km 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!]

Answer:

Time taken by the man to reach the market from home,

Time taken by the man to reach home from the market,

Total time taken in the whole journey = 30 + 20 = 50 min

Time = 50 min = 5/6h

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Speed of the man = 7.5 km

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min =

Net displacement = 2.5 - 1.25 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km