The average velocity for the first 4.9 hours of a trip is 27.2 km/hr east. It th

Question

The average velocity for the first 4.9 hours of a trip is 27.2 km/hr east. It then undergoes a second part of a trip where it has a different average velocity. If the whole trip takes 15.4 hours, what displacement, in km, does the object object travel for the second part of the trip if the average velocity for the entire trip is 16.6 km/hr east? If the answer is to the west, include a negative sign, but do not include a positive sign for the east.

The average speed for the first 3.5 hours of a trip is 33 km/hr.  It then undergoes a second part of a trip where it has a different average speed. If the whole trip takes 5 hours, what distance, in km, does the object travel for the second part of the trip if average speed for the entire trip is 45.9 km/hr?

Explanation / Answer

relation between spped and distance is Speed = distance/time

so Distance travelled in first 4.9 hrs = Vt

S1 = 27.2 * 4.9

S1 = 133.2 Km

time taken for travel in second part = 15.4-4.9 = 10.5 hrs

let V2 be the second speed, then Vavg = V1+ V2)/2

V2 = 2Vavg-V1

V2 = (2*16.6) - 27.2

V2 = 6kmph

so distance travelled S2 = 6 * 10.5 = 63 km

so total distance = 133.2+63 = 196.2 km

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relation between spped and distance is Speed = distance/time

so Distance travelled in first 3.5 hrs = Vt

S1 = 33*3.5

S1 = 115.5 Km

time taken for travel in second part = 5-3.5 = 1.5 hrs

let V2 be the second speed, then Vavg = V1+ V2)/2

V2 = 2Vavg-V1

V2 = (2*45.9) - 33.3

V2 = 58.5 kmph

so distance travelled S2 = 1.5*58.5 = 87.75 km

so total distance = 115.5 + 87.75 = 203.25 km

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