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dayshawn is traveling at 12 m/s away from the school. dayshaun's mom is looking for him because he is late coming home so she left the house and is driving toward the school at 5 m/s. The school is 6492 m from dayshauns house. at what position will dayshawn meet his mom and how long after she left the house did she find dayshaun? draw a motion map and solve. show your work

Respuesta :

Speed of Dayshawn travelling towards his home is 12 m/s

Speed of her mom towards his school is 5 m/s

They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home

Now let say they both meet after "t" time when they starts motion

so we can write the total distance between school and home as

[tex]d = v_1*t + v_2*t[/tex]

here d = 6492 m

[tex]v_1 = 12 m/s[/tex] = speed of dayshawn

[tex]v_2 = 5 m/s[/tex] = speed of his mom

now by solving the above equation

[tex]6492 = 12t + 5 t[/tex]

[tex]t = \frac{6492}{17}[/tex]

[tex]t = 381.9 s[/tex]

so they will meet after 381.9 s from start which will be 3.36 minutes from there start

Also at this time the distance covered by her mom will be

[tex]d_2 = v_2*t[/tex]

[tex]d_2 = 5* 381.9 = 1909.4 m[/tex]

so they will meet at distance 1909.4 m from their home

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