Question

An ant is moving along a rubber band at velocity v = 1cm/s. One end of the rubber band (the one from which the ant started) is fixed to a wall, the other (initially at distance L = 1m from the wall) is pulled at u = 1m/s. If the time taken by the out to reach the other end of the band is $(e^n-1)$ seconds, find n.

Answer 1

100

Answer 2

The problem describes an ant moving on a stretching rubber band. We need to find the time taken for the ant to reach the other end and then determine the value of 'n'.

Let's define the variables:

• $v$: Velocity of the ant relative to the rubber band = 1 cm/s = 0.01 m/s.
• $L$: Initial length of the rubber band = 1 m.
• $u$: Velocity at which the free end of the rubber band is pulled = 1 m/s.

Let $x(t)$ be the position of the ant from the fixed end at time $t$.
The length of the rubber band at time $t$ is $L(t) = L + ut$.

The velocity of the ant relative to the fixed end (ground) has two components:

1. Its own velocity relative to the rubber band, $v$.
2. The velocity of the point on the rubber band where the ant is located, due to the stretching of the band. A point at distance $x$ from the fixed end moves with a velocity proportional to its distance from the fixed end, so its velocity due to stretching is $\frac{x}{L(t)} u$.

Therefore, the total velocity of the ant relative to the fixed end is given by the differential equation: $\frac{dx}{dt} = v + \frac{x}{L(t)} u$ Substitute $L(t) = L + ut$: $\frac{dx}{dt} = v + \frac{x}{L + ut} u$ Rearrange it into a standard linear first-order differential equation form: $\frac{dx}{dt} - \left(\frac{u}{L + ut}\right) x = v$ This is of the form $\frac{dy}{dt} + P(t)y = Q(t)$, where $P(t) = -\frac{u}{L + ut}$ and $Q(t) = v$.

The integrating factor (IF) is $e^{\int P(t) dt}$: $\text{IF} = e^{\int -\frac{u}{L + ut} dt} = e^{-\ln(L + ut)} = e^{\ln((L + ut)^{-1})} = \frac{1}{L + ut}$ Multiply the differential equation by the integrating factor: $\frac{1}{L + ut} \frac{dx}{dt} - \frac{u}{(L + ut)^2} x = \frac{v}{L + ut}$ The left side is the derivative of $\left(\frac{x}{L + ut}\right)$ with respect to $t$: $\frac{d}{dt} \left(\frac{x}{L + ut}\right) = \frac{v}{L + ut}$ Now, integrate both sides with respect to $t$: $\frac{x(t)}{L + ut} = \int \frac{v}{L + ut} dt$ $\frac{x(t)}{L + ut} = \frac{v}{u} \ln(L + ut) + C$ To find the constant $C$, use the initial condition: at $t=0$, the ant starts from the fixed end, so $x(0) = 0$. $\frac{0}{L + u(0)} = \frac{v}{u} \ln(L + u(0)) + C$ $0 = \frac{v}{u} \ln(L) + C$ $C = -\frac{v}{u} \ln(L)$ Substitute $C$ back into the equation for $x(t)$: $\frac{x(t)}{L + ut} = \frac{v}{u} \ln(L + ut) - \frac{v}{u} \ln(L)$ $\frac{x(t)}{L + ut} = \frac{v}{u} \left[\ln(L + ut) - \ln(L)\right]$ Using logarithm properties, $\ln A - \ln B = \ln(A/B)$: $\frac{x(t)}{L + ut} = \frac{v}{u} \ln\left(\frac{L + ut}{L}\right)$ The ant reaches the other end when its position $x(t)$ is equal to the current length of the rubber band, $L(t) = L + ut$. Let $T$ be the time when this happens.
So, at $t=T$, $x(T) = L + uT$.
Substitute this into the equation: $\frac{L + uT}{L + uT} = \frac{v}{u} \ln\left(\frac{L + uT}{L}\right)$ $1 = \frac{v}{u} \ln\left(\frac{L + uT}{L}\right)$ Rearrange to solve for $T$: $\frac{u}{v} = \ln\left(\frac{L + uT}{L}\right)$ Exponentiate both sides: $e^{u/v} = \frac{L + uT}{L}$ $e^{u/v} = 1 + \frac{uT}{L}$ $e^{u/v} - 1 = \frac{uT}{L}$ $T = \frac{L}{u} (e^{u/v} - 1)$ Now, substitute the given values:
$L = 1 \text{ m}$
$u = 1 \text{ m/s}$
$v = 0.01 \text{ m/s}$ $T = \frac{1 \text{ m}}{1 \text{ m/s}} \left(e^{1 \text{ m/s} / 0.01 \text{ m/s}} - 1\right)$ $T = 1 \left(e^{100} - 1\right)$ $T = (e^{100} - 1) \text{ seconds}$ The problem states that the time taken by the ant to reach the other end is $(e^n - 1)$ seconds.
Comparing our result with the given form:
$(e^{100} - 1) = (e^n - 1)$
Therefore, $n = 100$.