Adriver 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).

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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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft