Posted by **thaokieu** on Sunday, January 30, 2011 at 8:02pm.

During the first 18 minutes of a 1.0-hour trip, a car has an average speed of 11 m/s. What must the average speed of the car be during the last 42 minutes of the trip be if the car is to have an average speed of 21 m/s for the entire trip?

- physics -
**Henry**, Tuesday, February 1, 2011 at 3:11pm
1hour = 3600s.

d = 21m/s * 3600s=75600m = t0tal distance to be traveled.

t1 = 18min.=18min * 60s/min = 1080s.

d1 = 11m/s * 1080s = 11880m. = Distanced traveled.

d2 = d - d1,

d2 = 75600 - 11880 = 63720m = distance remaining.

t2 = 42min = 42min * 60s/min = 2520s = Time remaining.

V(avg) = d2 / t2 = 63720 / 2520 = 25.3m/s = Average speed during t2.

CHECK:

V(avg) = d / (t1 + t2) = 75600 / 3600 = 21m/s = Average speed for the trip.

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