Q:

A tour bus normally leaves for its destination at 5:00p.m. for a 378 mile trip. This week however, the busleaves at 6:00 p.m. To arrive on time, the driver drives 9 miles per hour faster than normal What is the normalspeed of the bus?​

Accepted Solution

A:
Answer:54 mphStep-by-step explanation:we know D = RT, where D is distanceR is rate/speedT is timeNormally:D = RT378 = RTThis week:D = RT378 = (R+9)(T-1)These are the 2 equations we can write:Let's mutliply out the last equation:378 = (R+9)(T-1)378 = RT - R + 9T - 9387 = RT - R + 9Twe know RT = 378 and T = 378/R, we replace:387 = 378 - R + 9(378/R)9 = -R + 3402/RNow solving:[tex]9 = -R + \frac{3402}{R}\\R-\frac{3402}{R}+9=0\\R^2+9R-3402=0\\R=-63,54[/tex]R can't be negative, so we take R = 54 miles per hour