At time t=0 a car has a velocity of 16 m/s. It slows down with an acceleration g

Question

At time t=0 a car has a velocity of 16 m/s. It slows down with an acceleration given by -0.50t, in m/s^2 for t in seconds. It stops at t=

I have seen this problem answered wrong over and over, so since I finally have the answer, I will post it. The key is that the car's velocity needs to = 0. Integrate the acceleration. And at t=0 the velocity is 16, so add them together. V(t)=16-.25t^2. Think about it at t=0 the velocity is 16. Derivative of velocity is acceleration. That takes you back to -0.50t. Therefore 0=16-.25(t)^2 solve for t (-16/-.25)=t^2. Take the square root and the answer is t= 8.0s.  

Explanation / Answer

a = dv/dt => dv = (-0.5*t)*dt integrating both sides, we get; v = -0.5*t^2/2 + c at t=0, v = 16 m/s so, 16 = -0.5*0 +c => c = 16 so, equation of motion is; v = -0.5*t^2/2 + 16 now when it stops, v=0 => 0= -0.5*t^2/2 + 16 => t = 8 s

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