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Question

Heavy traffic reduced a drivers usual speed by 10mph, which lengthened her 200-mile trip by 1 hour. Find the driver's usual speed.

Answer 1

distance = rate x time

 

r is the "normal" rate of speed

t is the "normal" time

 

200 = r x t  = rt in regular traffic patterns

 

In heavy traffic,

 

200 = (r-10)(t+1)

 

rt = (r-10)(t+1)

 

rt = rt + r - 10t - 10

 

0 = r - 10t - 10

 

10t + 10 = r

10(t+1) = r

 

200 = r * t

200 = 10(t+1) t

 

200 = 10(t^2 + t)

20  = t^2 + t

0 = t^2 + t - 20

0 = (  t   +   5  )(  t  -     4     )

 

t + 5 = 0  ---> t = -5 which results in negatives times and rates

 

t - 4 = 0 ---> t = 4

 

She drives 200 miles in 4 hours, which is 200/4 = 50 mph.

 

In heavy traffic, she drives 200 miles at (50 -10) mph = 40 mph, 

which takes 200/40 = 5 hours, which is 1 hour longer.

 

Usual speed = 50 mph

Usual time = 4 hours