Question

In convey on a long straight level road, \[50\] identical cars are at rest in a queue at equal separation \[10m\] from each other as shown

Engine of a car provides a constant acceleration \[2\dfrac{m}{{{s}^{2}}}\] and brakes can provide a maximum deceleration\[4\dfrac{m}{{{s}^{2}}}\]. When an order is given to start the convey, the first car starts immediately and each subsequent car starts when its distance from a car that is immediately ahead becomes\[35m\]. Maximum speed limit on this road is\[72\dfrac{km}{h}\]. When an order is given to stop the convoy, the driver of the first car applies brakes immediately and driver of each subsequent car applies brakes with a certain time delay after noticing the brake light of the front car turned red.

During the time when motion is building up in the convoy, some of the cars are moving and the others are at rest. What is the average rate of change in length of the segment consisting of stationary cars?

Answer 1

Hint:First defined average velocity formula. Taking the given value and putting the second equation of motion will give the time interval for one car multiple with total car moving. Further distance covered by one car is multiplied by the total car in-between distance. After, the formula for finding the average rate of change in length of the segment.

Complete step-by-step solution:
Uniform motion:
An object can be termed as uniform motion if it covers equal distance in equal time intervals, however small these time intervals may be in the same fixed direction.
Average velocity:
For an object moving with not consistent velocity, average velocity is defined as the ratio of its total displacement to the total time interval in which that displacement occurs.
If \[{{s}_{1}}\] and \[{{s}_{2}}\] are the positions for an object at times \[{{t}_{1}}\] and\[{{t}_{2}}\], then the average velocity from time \[{{t}_{2}}\] and \[{{t}_{1}}\] is given by
\[{{v}_{av}}=\dfrac{{{s}_{2}}-{{s}_{1}}}{{{t}_{2}}-{{t}_{1}}}\]
At the start of convey one
Acceleration is given \[2\dfrac{m}{{{s}^{2}}}\]
The following car starts moving after the preceding car reaches a distance of \[35m\].
Time is taken to reach\[35m\], t have to be found
Equation of motion for the uniformly accelerated motion:
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Time is taken to reach\[35m\], t have to be found
\[35=\left( 0 \right)t+\dfrac{1}{2}\times 2{{t}^{2}}\]
\[{{t}^{2}}=35\]
\[t\simeq 6\sec \]
Each car start moving after \[6\] seconds
Initially, the length of the segments having stationary cars can be calculated by the space given between cars that will be \[49\]spaces and the distance between them is\[10m\].
Total \[50\] cars then, \[49\]spaces between them:
\[49\times 10=490\]
Thus, the distance becomes\[490m\].
After \[6\sec \]one car starts moving, which means the length of the segment between stationary cars decreases by \[10m\] after each\[6\sec \].
As there are \[50\] cars and the \[{{49}^{th}}\]car starts moving the distance between the stationary cars becomes\[0\].
It takes about \[49\times 6\]seconds for this
The Formula for Average acceleration:
\[{{v}_{av}}=\dfrac{{{s}_{2}}-{{s}_{1}}}{{{t}_{2}}-{{t}_{1}}}\]
\[=\dfrac{\Delta s}{\Delta t}\]

Where\[\Delta s\] is the distance and \[\Delta t\] is the average time
\[=\dfrac{490}{49\times 6}\]
\[=\dfrac{10}{6}\]
\[\simeq 2\dfrac{m}{s}\]
As the distance is decreasing with the car moving to the preceding car position, the Average rate of change of length of the segment will be decreasing too.
Thus, the final result is decreasing at\[2\dfrac{m}{s}\]

Note:The average velocity is always less than or equal to the average speed of an object. This can be seen by realizing that while distance is always strictly increasing, displacement can increase or decrease in magnitude as well as change direction. The average speed is a scalar quantity and not a vector quantity.