A group of tourists were supposed to cover an 18 km distance in a certain amount of time. However, they actually traveled at a speed which was .5 km/hour greater than what they had actually planned and thus arrived a half an hour early. What speed had the tourists planned to go at?

Respuesta :

Answer: The speed that they planed to go, was 4 km/h.

Step-by-step explanation:

Let tourists planned to go at the speed of x km/h,

After increasing the speed by 0.5 km/h,

Their new speed = ( x + 0.5 ) km/h,

Total distance = 18 km,

According to the question,

The time taken at the speed of x - time taken at the speed of (x+0.5) = 1/2 hours,

[tex]\implies \frac{18}{x}-\frac{18}{x+0.5}=\frac{1}{2}[/tex]

( Because, Time = distance / speed ),

[tex]\frac{18(x+0.5)-18x}{x^2+0.5x}=\frac{1}{2}[/tex]

[tex] \frac{9}{x^2+0.5x}=\frac{1}{2}[/tex]

[tex]18 = x^2+0.5x[/tex]

[tex]180 = 10x^2 + 5x[/tex]

[tex] 10x^2+5x-180=0[/tex]

[tex] 5x^2+x-36=0[/tex]

[tex]2x^2 + 9x - 8x - 36 = 0[/tex]

[tex] x(2x + 9) - 4(2x + 9) = 0[/tex]

[tex](x - 4)(2x + 9) = 0[/tex]

⇒ x - 4 = 0 or 2x + 9 = 0  

x = 4 or 2x = -9  

x = 4 or x = -9/2 = -4.5 (not possible)

Thus, the speed that they planed to go = x km/h = 4 km/h

The Planned Speed of the Tortoise is 4 km/h.

How do you calculate the planned speed?

Given that, the total traveling distance is 18 km.

Let us consider that, the Planned Speed is y km/hr. So the time taken to travel the distance with Planned Speed is,

[tex]Time = \dfrac {18}{y}[/tex]

Now the Actual Speed is 0.5 km/hr more than the Planned Speed. So

Actual Speed = [tex](y+0.5) \rm km/hr[/tex]

Time taken to travel the distance with Actual Speed is,

[tex]Time (Actual Speed) = \dfrac {18}{y+0.5}[/tex]

Given that, the tortoise arrived half hour before when they traveled with the Actual Speed. So, the difference between the time is given as,

[tex]\dfrac {18}{y} - \dfrac {18}{y+0.5} = \dfrac {1}{2}[/tex]

Simplifying the above equation as given below.

[tex]18\times 2 \; (\dfrac{1}{y} - \dfrac{1}{y+0.5}) = 1[/tex]

[tex]36 \; (y+0.5 -y) = 1\times y\times (y+0.5)[/tex]

[tex]36\times 0.5 = y^2 + 0.5y[/tex]

[tex]y^2 +0.5y=18[/tex]

[tex]y^2+0.5y-18 =0[/tex]

The above equation can be written as given below.

[tex]2y^2 +y-36=0[/tex]

[tex]2y^2 +9y-8y-36=0\\[/tex]

[tex]y(2y+9)-4 (2y+9)=0[/tex]

[tex](y-4)(2y+9)=0[/tex]

[tex]y=4\; or\; y= \dfrac{-9}{2}[/tex]

Thus the Planned Speed is 4 km/hr.

For more details about the speed, follow the link given below.https://brainly.com/question/1889005.