yesterday i ran across a problem that said "a car is at rest and accelerates at 2 meters per second^2. how long does it take the care to go 30 meters?" The book used the physics formula and got 5.48 seconds, but i wanted to use my calculus skillz to figure it out. so i said that f''(x) is the acceleration function and f''(x)= 2x with x being time in seconds. Then I integrated that. f'(x)= x^2 (that being the velocity formula in meters per second and C=0 in this integration since the car is at the starting point of 0). Then I integrated again to get f(x)=(x^3)/3 (this being the position function for the car in meters). so when f(x)=30 meters you should have the right answer right? I plugged it into my calculator and found that when x=4.48 seconds then f(x)= 30 meters.... 4.48 was slightly different from the book's 5.48. so what did i do wrong? are derivatives/integration even a usable option in this scenario?
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The acceleration is constant, not a function of time:
f"(t) = 2 = a
f'(t) = ∫a*dt = 2*t = v
f(t) = ∫v*dt = ½*2*t² = 1.0t² = x
For x = 30,
t = √30 = 5.48
f"(t) = 2 = a
f'(t) = ∫a*dt = 2*t = v
f(t) = ∫v*dt = ½*2*t² = 1.0t² = x
For x = 30,
t = √30 = 5.48