Question

A brand-new engineering hire is late for her first product development team meeting. She gets out of her car and starts running 8 mph. It is exactly 7:58 a.m., and the meeting starts at exactly 8:00 a.m. Her meeting is 500 yd away. Will she make it on time to the meeting? If so, with how much time to spare? If not, how late will she be?

Answer 1

Obtain the relationship between yard and feet as follows:

1 yard = 3 feet

 

Now, convert the distance to be covered in terms of feet as follows

(500 yd) × (3 ft/1 yd) = 1500 ft

 

Use the below conversion factor from table-“Certain derived units in the USCS”

1 mile = 5280 feet

 

Convert the distance to be covered in miles as follows:

(1500 ft) × (1 mi/5280 ft) = 0.284 mi

 

The speed of the hire is 8 mph.

 

Convert the speed into mi/s as follows

(8 mi/hr) × (1 hr/3600 s) = 0.0022 mi/s

 

Obtain the time required to cover a distance of 0.284 mi at the speed of 0.0022 mi/s

 

Time required t = (0.284 mi) × (1/0.0022 mi/s)

= 129 s

 

Therefore, the hire will not make it in time and will be late by 9 seconds as the distance has to be covered in 120 seconds.

Therefore, the hire will not make it in time and will be late by 9 seconds as the distance has to be covered in 120 seconds.