Atruck enters a stretch of road that drops 4 meters in elevation for every 100 meters along the length of the road. the road is at 1,300 meters elevation where the truck entered, and the truck is traveling at 16 meters per second along the road. what is the elevation of the road, in meters, at the point where the truck passes t seconds after entering the road?
a. 1,300 - 0.04t
b. 1,300 - 0.64t
c. 1,300 - 4t
d. 1,300 - 16t

The elevation of the road at the point where the truck passes is 1,300 - 0.64t.

The given parameters;

The elevation of the road at the point where the truck passes is calculated as follows;

h = y + Δy(vt)

Thus, the elevation of the road at the point where the truck passes is 1,300 - 0.64t.

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  B.  1,300 - 0.64t

Step-by-step explanation:

The change in elevation is -4/100 = -0.04 of the distance traveled. When traveling at 16 m/s, the distance traveled is 16t. -0.04 that amount is ...

  -0.04 × 16t = -0.64t

The elevation starts at 1300 m and changes by that amount as a function of time, so can be described by ...

  1300 -0.64t . . . . . matches choice B

eve and marks ratios are equivalent

step-by-step explanation:

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol