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

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MathPhys

Answer:

9 ft/s²

Explanation:

1)

d²s/dt² = -k

Integrate once:

ds/dt = -kt + C

At t=0, ds/dt = 66:

66 = -k(0) + C

C = 66

ds/dt = -kt + 66

Integrate again:

s = -½ kt² + 66t + D

At t=0, s = 0:

0 = -½ k (0)² + 66 (0) + D

D = 0

s = -½ kt² + 66t

2)

Setting ds/dt = 0:

0 = -kt + 66

t = 66/k

3)

Setting s = 242:

242 = -½ kt² + 66t

242 = -½ k (66/k)² + 66 (66/k)

242 = -2178/k + 4356/k

242 = 2178/k

k = 9

The driver must decelerate at 9 ft/s².

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