Answer to Question #282175 in Physics for suz

Question

It is now 8:59 am, but when the bell rings at 9:00 am Suzette will be late for Science class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she covers its 48.0 m length at an average speed of 1.20 m/s. The final hallway is empty, and Suzette spring its 60.0 m length at a speed of 5.00 m/s Determine if Suzette makes it to class on time by finding how many seconds she arrives before or after the bell.

Answer 1

To find the answer, simply find the time required to run across each hallway and sum it up! Remember that time is distance divided by speed.

1) 35.0 m at a speed of 3.50 m/s:

$$t_1=35/3.5=10\text{ s}.$$

2) 48.0 m at an average speed of 1.20 m/s:

$$t_2=48/1.2=40\text{ s}.$$

3) 60.0 m at a speed of 5.00 m/s:

$$t_3=35/3.5=12\text{ s}.$$

The total time, therefore, is

$$t=t_1+t_2+t_3=62\text{ s}.$$

Suzette will be 2 s late.