Question

An athlete us jumping. However, everytime she jumps she gets a bit more tired, and every jump goes 1/2 as far as her prior jump. Now, for her very first jump, she goes 1/2 of a foot. On her second jump, she goes 1/4 of a foot, and so on and so forth. How many jumps does it take to get to travel 1 foot

Answer 1

She can never travel a distance of 1 foot for the given condition.

Explanation:

In every jump, she goes 1/2 as far as her prior jump.

In the 1st jump, she covered a distance, d_1= \frac 1 2 foot.

The distance covered in the 2nd jump

`d_2= \frac 12 * d_1 = = \frac 12 * \frac 12= \frac 1 4` (given)

So, the distanve covered in the 3rd jump

`=\frac 12 * d_2= \left(\frac 1 2\right )^2* \frac 1 2= \left(\frac 1 2\right)^(3-1)* \frac 1 2.`

Similarly, the distance covered in the
`r^(th)` jump

`=\left(\frac 12 \right)^(r-1)* \frac 1 2.`

Assuming she requires
`n` jumps to travel 1 foot of distance.

So, the sum of all the distances covered in n jumps = 1 foot

`\Rightarrow \frac 1 2 + \frac 1 4 + \cdots + \left(\frac 12 \right)^(r-1)* \frac 1 2+\cdots+ \left(\frac 12 \right)^(n-1)* \frac 1 2=1`

`\Rightarrow \frac 1 2 + \left(\frac 12 \right)^2 + \cdots + \left(\frac 12 \right)^(r)+\cdots+ \left(\frac 12 \right)^(n)=1`

This a geometric progression, G.P., of
`n` terms having common ration, r= 1/2 and the first term
`a_1= 1/2`.

As sum of all the n terms of G.P
`=\frac {a_1(r^n-1)}{r-1}`,

`\Rightarrow \frac {1/2((1/2)^n-1)}{1/2-1}=1`

`\Rightarrow - \left(\left(\frac 1 2\right)^n-1\right)=1`

`\Rightarrow \left(\frac 1 2\right)^n-1=-1`

`\Rightarrow \left(\frac 1 2\right)^n=-1+1=0`

`\Rightarrow \frac {1}{2^n}=0`

This is only possible, mathematically, when
`n\rightarrow \infty`, But in real life situation reaching infinity is not possible.

Hence, she can never travel a distance of 1 foot for the given condition.