June 8, 2008

Q. 1.3 -- "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$:

A car starts moving recilinearly, first with acceleration $\omega$= 5 $m/s^2$(the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\omega$, comes to a stop. The total time of motion equals $\tau$= 25 s. The average velocity during this time is equal to $\emph{v}$= 72 km per hour. How long does the car move uniformly?

$\textbf{Answer}$:

This question can be answered by calculating the distance using two methods:

  1. Using average speed for the entire journey.
  2. Using the distance covered by the car during different sections of its journey.

Calculate distance covered using average speed:
Total distance covered during journey = $\emph{v} \tau$

Calculate distance covered using sectional speeds:
The car spends equal amount of time during acceleration and deceleration. Lets call it $t$

Distance covered during acceleration phase = $0t + \frac{1}{2}\omega t^2$= $0.5\omega t^2$

The same distance is also covered during deceleration = $0.5\omega t^2$

Also, the velocity of car after acceleration phase is = $0 + \omega t$= $\omega t$

The car travels for total of $\tau - 2t$time at this velocity. So the distance covered at constant speed is $\omega t(\tau - 2t)$.

Distance calculated using both methods must be equal and hence,

$\omega t (\tau - 2t) + 0.5 \omega t^2 \times 2$= $\emph{v} \tau$

$\Rightarrow \omega t^2 - \omega t\tau + v\tau = 0$

Solving for $t$, we get:

\begin{displaymath}t = \frac {\omega\tau \pm \sqrt{\omega^2\tau^2 - 4 \omega v \tau}}{2\omega} \end{displaymath}


\begin{displaymath}\Rightarrow t = \frac {\omega\tau \pm \omega\tau \sqrt {1 - 4v/(\omega\tau)}}{2\omega} \end{displaymath}


\begin{displaymath}\Rightarrow 2t = \tau \pm \tau \sqrt {1 - 4v/\omega\tau} \end{displaymath}


\begin{displaymath}\Rightarrow \tau - 2t = \pm \tau \sqrt {1 - 4v/\omega\tau} \end{displaymath}

But since the car traveled for a finite time at constant speed, . Thus, the time for which the car travels at constant speed is:

\begin{displaymath}\tau - 2t = \tau \sqrt {1-4v/\omega\tau}\end{displaymath}

Using the value provided,

\begin{displaymath}\tau - 2t = 25 \times \sqrt {1- 4 \times 20 / (5 * 25)} = 25 \times \sqrt {9/25} = \frac {25 \times 3}{5} = 15 \end{displaymath}

Posted by Apurva Sharan at 5:44 PM
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