Antiderivatives: A car is traveling at 50mi/h when the brakes are fully applied, producing a constant deceleration of 38ft/s2. What is the distance covered before the car comes to a stop?Since the car is decelerating, our value will be negative. v ′ ( t ) = a ( t ) = − 38 so v ( t ) = − 38 x = s ( t ) − 38 x 2 / ( 2 ).After this point I am not sure what I am supposed to do?

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Antiderivatives: A car is traveling at 50mi/h when the brakes are fully applied, producing a constant deceleration of 38ft/s2. What is the distance covered before the car comes to a stop?Since the car is decelerating, our value will be negative.      v    ′    (  t  )  =  a  (  t  )  =  −  38 so   v  (  t  )  =  −  38  x  =  s  (  t  )  −  38      x    2        /    (  2  ).After this point I am not sure what I am supposed to do?

Josie123 · Accepted Answer

Step 1Well let us first attempt to get our units the same. For miles/hour we can rewrite this as       220    3   feet per second.We are given an acceleration, an initial velocity, and are asked for a distance.Step 2            d      v              d      t        =            d      v              d      x                  d      x              d      t        =  a    ⟹    v  d  v  =  a  d  xIntegrating from initial velocity given to 0, and using the given acceleration, we find:  0  −      1    2    (      220    3        )    2    =  (  −  38  )  Δ  xOr,   Δ  x  =  70.7602  f  t

Antiderivatives: A car is traveling at 50mi/h when the brakes are fully applied, producing a constan

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