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Each second a rabbit moves one-half the remaining distance from its nose to a head of lettuce. Does the rabbit ever get to the lettuce? What is the limiting value of the rabbit’s average velocity? Draw graphs showing the rabbit’s velocity and position as time increases.

Each second a rabbit moves one-half the remaining distance from its nose to a head of lettuce. Does the rabbit ever get to the lettuce? What is the limiting value of the rabbit’s average velocity? Draw graphs showing the rabbit’s velocity and position as time increases.

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Yes, the rabbit will get to the lettuce. To understand it better, let us assume that initially, the distance between the rabbit and the lettuce (x) is s . Therefore the distance traveled by the rabbit in 1st second is s/2 . The remaining distance to the lettuce s/2 is and the distance traveled by the rabbit now will be exactly one-half this value i.e. s/4 .
Therefore, the distance between the rabbit and lettuce falls every second as:
x = s \ (at \ t = 0\ sec), x = \frac{s}{2} (at\ t\ = \1\sec), x = \frac{s}{4} , x = \frac{x}{4} (at \ t = \3 \ sec)............
After some time, the distance between the rabbit and the lettuce will be so small that the rabbit will almost reach its destination. The limiting value in this case will occur for the change in distance\Delta x \rightarrow 0 .
The table below shows the limiting process, in which the initial point x1 (m) of the rabbit (measured from the starting point) tends to reach the final point x2 (m) (measured from starting point), over each time interval of 1 sec.



233-1807_13.PNG

Therefore the limiting value of average velocity will be closer to s /128 for a particular value of s .
It is important to note that the value of initial point in the table above is measured from the starting point. After one second, the rabbit has travelled distance s/2 , therefore he is that much distance away from the initial point. After another second, the rabbit has travelled a distance of s / 4 , therefore the rabbit s/2 +s/4 is meters away from the starting point i.e. 3s/4. This is repeated for every second and the limiting condition is approached.
The position (x) of the rabbit from the starting point after t seconds is given as:
x = s-s \left ( \frac{1}{2} \right )^{t}
One can match the value x1 (m) of in the table above with the values of (x) for . t = 1,2,3....
The graph below shows the position of rabbit from the starting position after t seconds:
233-568_2.png
It can be seen that after t = 1 s, the rabbit s/2 is or 0.5s away from the initial point. After another 1 second or t = 2 , the rabbit is 3s /4 or 0.7s away from the starting point and so on.

The velocity of the rabbit at any after ant time is the differentiation of the position (x) , and is given as:
velocity = \frac{dx}{dt}
= \frac{d\left ( s - s\left ( \frac{1}{2} \right )^{t} \right )}{dt}
= -\left (\frac{1}{2} \right )^{t} In \left ( \frac{1}{2} \right )
Here, we have used the formula of differentiation i.e. \frac{d(a^{x})}{dx} = a^{x} In \ a..
Therefore the velocity of the rabbit, after substituting the value of In \left ( \frac{1}{2} \right ) as -0.69, is:
velocity = (-0.69) \left ( -\frac{1}{2} \right )^{t}
= 0.69 \left ( \frac{t}{2} \right )^{t}
The plot the velocity of the rabbit is shown below:

233-319_1111.png
It can be seen from the graph above, that at t = 0, the velocity of the rabbit is maximum i.e. 0.69s. After another one second, the distance travelled by the rabbit is smaller and therefore its velocity has fallen down. The velocity continues to fall over time thereby accounting for the smaller distance travelled by the rabbit each second.

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