Question

A traffic officer receives complaints of often traffic jams at the traffic signals on the main street of a busy market. He studied the traffic pattern to simplify calculations and made reasonable assumptions that all the vehicles are identical in size and move with identical speed. He also notices that when a signal turns green, all the vehicles do not begin to move at the same time; but vehicles standing adjacent to traffic signals begin to move first after a small-time delay, then next after the same time delay, and so on. At present, durations of red and the green signals are equal an average speed of traffic advancement is ${v_1} = 1.5m{s^{ - 1}}$ If he orders to make $\eta = 2$ times the duration of green signal and leaves the duration of the red signal unchanged, what would be the average speed ${v_2}$ of the traffic advancement?

Answer 1

Hint: Speed of an object shows how fast the object is moving. If the speed of an object is high it means the object is moving fast. Velocity is the rate of change of position of an object with time in a given direction. To solve the given problem, write down the provided physical quantities and then carefully apply the formula for average speed on them to find the required correct answer.

Formula used: ${\text{Average speed}} = \dfrac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$

Complete step by step answer:
For an object moving with variable speed, the average speed is the total distance travelled by the object divided by the total time taken to cover that distance i.e. ${\text{Average speed}} = \dfrac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$

The duration of traffic lights controls how long the cars will move and how long they will remain constant which controls the average speed of the cars. According to the question, at ${v_1} = \dfrac{s}{{t + t}}$ -- (i) Where, $s$ is the distance and each $t$ represents the duration of red and green signals.
We know that the cars will move only when they see green signals. But we need to consider the time of both the red and green signal’s duration.
From equation $(i)$, $s = 2t{v_1}$
But now the duration of the green signal becomes $\eta t$. So, distance covered will be $s = 2\eta t{v_1}$ Total duration of time which is duration of green and red signal respectively will be, $T = \eta t + t$
Thus,
Average velocity ${v_2} = \dfrac{{2\eta t{v_1}}}{{(\eta + 1)t}} = \dfrac{{2\eta {v_1}}}{{(\eta + 1)}}$
Since, it is given that in order to improve the situation, the traffic officer orders to make the duration of green signal $\eta = 2$ and ${v_1} = 1.5m{s^{ - 1}}$.
${v_2} = \dfrac{{2\eta {v_1}}}{{(\eta + 1)}}$
$ \Rightarrow {v_2} = \dfrac{{2(2)(1.5)}}{{(2 + 1)}} = 2m/s$

Hence, the average speed ${v_2}$ of the traffic advancement would be $2m/s$.

Note: Average velocity of an object is defined as the ratio of its total displacement to the total time interval in which the displacement occurs i.e. ${\text{Average velocity = }}\dfrac{{{\text{Total displacement}}}}{{{\text{Total time }}}}$
Speed is a scalar quantity, so average speed can only be positive and zero. While velocity is a vector quantity, i.e. it has direction associated with it as well as magnitude. So it can be positive, negative or zero.