Respuesta :
Answer:
[tex]d=2.4\ miles[/tex]
Explanation:
Given:
- average walking speed, [tex]v_w=3\ mph[/tex]
- average biking speed, [tex]v_b=12\ mph[/tex]
According to given condition:
[tex]t_w=t_b+\frac{36}{60}[/tex]
where:
[tex]t_w=[/tex] time taken to reach the building by walking
[tex]t_b=[/tex] time taken to reach the building by biking
We know that,
[tex]\rm time=\frac{distance}{speed}[/tex]
so,
[tex]\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}[/tex]
[tex]\frac{d}{3}=\frac{d}{12} +\frac{3}{5}[/tex]
[tex]d=2.4\ miles[/tex]
Answer:
The distance from her apartment to the classroom building is 2.4 miles.
Explanation:
Given that,
Time = 36 min
Walking average speed of her = 3 m/h
biking average speed of her = 12 m/h
If she takes n minutes to ride, then if the distance is d miles,
We need to calculate the distance
Using formula of time
[tex]t=t_{1}+t_{2}[/tex]
[tex]t=\dfrac{d}{v_{1}}+\dfrac{d}{v_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{36}{60}=\dfrac{d}{3}-\dfrac{d}{12}[/tex]
[tex]d=\dfrac{36\times12}{60\times3}[/tex]
[tex]d=2.4\ miles[/tex]
Hence, The distance from her apartment to the classroom building is 2.4 miles.
