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It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be latefor French class for the third time this week. She must get from one side of theschool to the other by hurrying down three different hallways. She runs downthe first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The secondhallway is filled with students, and she covers its 48.0-m length at an averagespeed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0-mlength at a speed of 5.00 m/s. a) Does Suzette make it to class on time or doesshe get detention for being late again? b) Draw a distance vs. time graph ofthe situation.

Respuesta :

Answer:

a) She get detention for being late again

Explanation:

First, we need to identify how much time does she take on each hallway.

With the distance and the speed, we can calculate the time as:

t = distance/speed

So, for each hallway, we get:

First hallway:

distance = 35 m

speed = 3.5 m/s

time = 35/3.5 = 10s

Second Hallway

distance = 48 m

speed = 1.2 m/s

time = 40s

Third Hallway

distance = 60 m

speed = 5 m/s

time = 60/5 = 12 s

Therefore, the total time that she takes was

10s + 40s + 12s = 62s

Since she takes more than 60 seconds, she will be late again.

Finally, we know that she takes 10s to run a distance of 35m, then another 40s to run a distance of 48 m, and another 12s to run a distance of 60 m. Therefore, the distance vs. time graph for this situation is

Ver imagen CaseT153835

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