Vehicle A is 819.017 feet behind vehicle B. Vehicle A is traveling at 93.602fps and vehicle B is traveling at 44fps. How long until vehicle A catches vehicle B?
so the 2nd vehicle must make up the distance of 819.017 ft
let the time taken to catch up be t seconds
93.602t - 44t = 819.017
solve for t
in time t
vehicle A travels distance d
vehicle B travels distance (d-819.017)
93.602 t = d
44 t = d-819.017
so
44 t = 93.602 t -819.017
To find out how long it will take vehicle A to catch up with vehicle B, we need to determine how long it will take for their distances to be equal.
Let's assume that both vehicles have been traveling for t seconds when vehicle A catches up to vehicle B.
So, the distance traveled by vehicle A in t seconds can be calculated using the formula: Distance = Speed x Time
Therefore, the distance traveled by vehicle A is: Distance A = (93.602 fps) x t
Similarly, the distance traveled by vehicle B in t seconds is: Distance B = (44 fps) x t
As per the given information, Vehicle A is 819.017 feet behind Vehicle B.
Thus, we can set up an equation: Distance A - Distance B = 819.017
Substituting the distances we calculated earlier, we get: (93.602t) - (44t) = 819.017
To solve this equation, we combine like terms: 93.602t - 44t = 819.017
Simplifying further, 49.602t = 819.017
To isolate t, we divide both sides by 49.602: t = 819.017 / 49.602
Calculating this result, t ≈ 16.50 seconds. Therefore, it will take approximately 16.50 seconds for vehicle A to catch up with vehicle B.