# A driver driving along a highway at a steady 45 mph â€‹(66 â€‹ft/sec) sees an accident ahead and slams on the brakes. What constant deceleration is required to stop the car in 242 â€‹ft? To findâ€‹ out, carry out the following steps. â€‹(1) Solve the following initial value problem. StartFraction d squared s Over dt squared EndFraction equals negative k â€‹(k constant), with StartFraction ds Over dt EndFraction equals66 and sequals0 when tequals0 â€‹(2) Find the value of t that makes StartFraction ds Over dt

1 answer

###### Question:

A driver driving along a highway at a steady 45 mph â€‹(66 â€‹ft/sec) sees an accident ahead and slams on the brakes. What constant deceleration is required to stop the car in 242 â€‹ft? To findâ€‹ out, carry out the following steps. â€‹(1) Solve the following initial value problem. StartFraction d squared s Over dt squared EndFraction equals negative k â€‹(k constant), with StartFraction ds Over dt EndFraction equals66 and sequals0 when tequals0 â€‹(2) Find the value of t that makes StartFraction ds Over dt EndFraction equals0. â€‹(The answer will involveâ€‹ k.) â€‹(3) Find the value of k that makes sequals242 for the value of t found in the stepâ€‹ (2).

## Answers

Answer:9 ft/sÂ²Explanation:1)dÂ²s/dtÂ² = -kIntegrate once:ds/dt = -kt + CAt t=0, ds/dt = 66:66 = -k(0) + CC = 66ds/dt = -kt + 66Integrate again:s = -Â½ ktÂ² + 66t + DAt t=0, s = 0:0 = -Â½ k (0)Â² + 66 (0) + DD = 0s = -Â½ ktÂ² + 66t2)Setting ds/dt = 0:0 = -kt + 66t = 66/k3)Setting s = 242:242 = -Â½ ktÂ² + 66t242 = -Â½ k (66/k)Â² + 66 (66/k)242 = -2178/k + 4356/k242 = 2178/kk = 9The driver must decelerate at 9 ft/sÂ².

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