Question

A turtle lives in a garden and a hedgehog lives in the woods. They leave their homes at the same time walk toward each other and meet in $5hours$ . The turtle walks $10metersperhour$ slower than the hedgehog. If the turtle had left home $4\frac{1}{2}hours$ earlier than the hedgehog had left his home, the two would meet $150m$ from the woods. Find the distance between the garden and the woods.

Answer 1

Step-1: Find the expression:

Let the distance between the garden and the woods $=d$.

The velocity of the hedgehog $=v$

The time they take to meet each other $=t$

$10m=0.01km$

$150m=0.150km$

According to the statements:

The velocity of the Turtle $=v-0.01$

$d=5(v-0.01)+5v$ $...\left[1\right]$

$(v-0.01)(t+4.5)=d-0.15$ $...\left[2\right]$

$vt=0.15$ $...\left[3\right]$

Step-2: Find the expression in one variable:

Equation $\left[1\right]$ can be written as:

$d=5v-0.05+5v$

$\Rightarrow$ $d=10v-0.05$ $...\left[4\right]$

Equation $\left[3\right]$ can be written as:

$t=\frac{0.15}{v}$ $...\left[5\right]$

Equation $\left[2\right]$ can be written as:

$vt+4.5v-0.01t-0.045=d-0.15$ $...\left[6\right]$

Put the value of $d\&t$ from $\left[4\right]\&\left[5\right]$ in $\left[6\right]$

$0.15+4.5v-0.01\times \frac{0.15}{v}-0.045=10v-0.05-0.15$

$0.105+4.5v-\frac{0.0015}{v}=10v-0.2$

$10v-4.5v+\frac{0.0015}{v}-0.105-0.2=0$

$5.5v+\frac{0.0015}{v}-0.305=0$

$5.5v^2-0.305v+0.0015=0$ $...\left[7\right]$

Step-3: Solving the quadratic equation:

Use the Sridharacharya formula to solve for $v$.

Sridharacharya formula to find the value of $x$ for quadratic equation $a{x}^2+bx+c=0$, $...\left[8\right]$

where $a,b\&c\in R$ and $a\ne 0$ is given by

$x=\frac{\left(-b\pm \sqrt{b^2-4ac}\right)}{2a}$ $...\left[9\right]$

Compare $\left[7\right]\&\left[8\right]$. we get

$a=5.5,b=-0.305\&c=0.0015$

Put the above values in $\left[9\right]$, we get

$v=\frac{0.305\pm \sqrt{{\left(0.305\right)}^2-4\times 5.5\times 0.0015}}{2\times 5.5}$

$\Rightarrow$ $v=\frac{0.305\pm \sqrt{0.093025-0.033}}{11}$

$\Rightarrow$ $v=\frac{0.305\pm \sqrt{0.060025}}{11}$

$\Rightarrow$ $v=\frac{0.305\pm 0.245}{11}$

On taking positive sign

$v=\frac{0.305+0.245}{11}$

$\Rightarrow$ $v=\frac{0.55}{11}$

$\Rightarrow$ $v=0.05$

On taking negative sign.

$v=\frac{0.305-0.245}{11}$

$\Rightarrow$ $v=\frac{0.305-0.245}{11}$

$\Rightarrow$ $v=\frac{0.06}{11}$

$\Rightarrow$ $v=0.00545$

Hence $0.05\&0.00545$ are two values of $v$.

Step-4: Find the value of $d$:

Put $v=0.00545$ in $\left[4\right]$

$\Rightarrow$ $d=10\times 0.00545-0.05$

$\Rightarrow$ $d=0.0545-0.05$

$\Rightarrow$ $d=0.0045$

$\Rightarrow$ $d>0.15$ $d\ne 0.0045$

Put $v=0.05$ in $\left[4\right]$.

$\Rightarrow$ $d=10\times 0.05-0.05$

$\Rightarrow$ $d=0.5-0.05$

$\Rightarrow$ $d=0.45$

Hence, the distance between the garden and woods is $0.45km$ or $450m$.