Question

Un ciclista A marcha a 24 Km/h y un ciclista B se desplaza a 6,8 m/s, Ambos están ubicados en un camino que tiene 4 Km de largo, pero el ciclista A está ubicado en el inicio del camino y el ciclista B está a 500 metros después. Conteste: a) Cuál de los dos ciclistas se desplaza a mayor velocidad b) Cuánto tiempo demora cada ciclista en llegar al extremo final del camino

Answer 1

(a) Cyclist A travels faster.

(b) The time taken by cyclists A and B is 10 minutes and 8.58 minutes respectively.

Step-by-step explanation:

Given that the speed of cyclist A= 24 km/h

As 1 km = 1000 m and 1 hour= 3600 seconds.

So,
`24 \frac {km}{h}=24* (1000 m)/(3600 s)=(20)/(3)\frac {m}{s}=6.67 m/s`

and the speed of cyclist B = 6.8 m/s.

The length of the road = 4 km=4000 m.

(a) Speed of cyclist A is 6.67 m/s which is more than the 6.8 m/s.

Hence, cyclist A travels faster.

(b) If d be the distance to be traveled and V be the speed, than the time, t, required is,

`t=(d)/(V)\cdots(i)`

For cyclist A:

As A is located at the beginning of the road, he has to cover full length of the road.

So, the distance to be covered by cyclist A =4 km.

Hence, from equation (i), the time taken by cyclist A is

`t=(4)/(24)=(1)/(6)` hour

`=(1)/(6)* {60}=10` minutes. [ as 1 hour = 60 minumes]

So, time taken by cyclist A is 10 minutes.

For cyclist B:

As B is located 500 meters later from the beginning of the road, he has to cover the remaining length of the road.

So, the distance to be covered by cyclist B =4000 m-500 m= 3500 m.

Hence, from equation (i), the time taken by cyclist B is

`t=(3500)/(6.8)=514.71` seconds

As 60 seconds= 1 minute.

So, t= 514.71/60 minutes = 8.58 minutes.

`=(1)/(6)* {60}=10` minutes. [ as 1 hour = 60 minumes]

So, the time taken by cyclist B is 8.58 minutes.