Math, asked by jjk20612251, 11 days ago

3) Shayna left the airport and drove toward the capital at an average speed of 52 km/h. Some time later Mei left driving in the same direction but at an average speed of 65 km/h. After driving for four hours Mei caught up with Shayna. Find the number of hours Shayna drove before Mei caught up.​

Answers

Answered by REONICKSTAR
1

Given data : Shayna left the airport and drove toward the capital at an average speed of 52 km/hr. Some time later Mei left driving in the same direction but at an average speed of 65 km/hr. After driving for four hours Mei caught up with Shayna.

To find : The number of hours Shayna drove before Mei caught up.

Solution : According to given data here we know that, Shanaya and Mei covered equal distance.

Let, distance covered by Shanaya and Mei be x km.

  • speed of Shanaya = 52 km/hr
  • time taken by Shanaya = ?
  • speed of Mei = 65 km/hr
  • time taken by Mei = 4 hour

Now,

⟹ SPEED = DISTANCE/TIME

⟹ DISTANCE = SPEED * TIME ----{1}

Now,

⟹ Distance covered by Shanaya = Distance covered by Mei

⟹ 52 * Time = 65 * 4 [ from {1} ]

⟹ 52 * Time = 260

⟹ Time = 260/52

⟹ Time = 5 hour

Answer : Shanaya travel 5 hour when caught by Mei.

{More info : conversion hour to second

  • Time = 5 hour
  • Time = {5 * 60 * 60} sec
  • Time = {5 * 3600}
  • Time = 18000 sec}

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