Under the assumption of <em>uniform</em> motion, the traveller will travel a distance of 20 feet before slowing down the vehicle to less than 40 miles per second.
<h3>How to calculate the distance taken by traveler before braking</h3>
According to this question, the vehicle travels at <em>constant</em> velocity before the traveler hit the brakes. The distance <em>traveled</em> (s), in feet, is equal to the product of velocity (v), in feet per second, and time (t), in seconds:
s = v · t (1)
If we know that v = 40 ft/s and t = 0.5 s, then the traveled distance of the vehicle is:
s = (40 ft/s) · (0.5 s)
s = 20 ft
Under the assumption of <em>uniform</em> motion, the traveller will travel a distance of 20 feet before slowing down the vehicle to less than 40 miles per second.
To learn more on uniform motion: brainly.com/question/118814
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V=lwh1/2
V=(24•20•30)1/2
V=17280•1/2
V=8,640
Hope this helps!
60-14 , 25x46=1150. Thats how you solve it
Answer: 14/15 of a meter
Step-by-step explanation:
5 and 3 LCM is 15.
3/5 x 3 + 1/3 x 5= 9/15 + 5/ 15 = 14/15
I think the answer's C: $7000/$8000
That's a surprisingly easy question for high school
