Question

You drive on Interstate 10 from San Antonio to Houston, half the time at 51 km/h and the other half at 71 km/h. On the way back you travel half the distance at 51 km/h and the other half at 71 km/h.

What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip?

Answer 1

Part a)

`v_(avg) = 61 km/h`

Part b)

`v_(avg) = 59.4 km/h`

Part c)

`v_(avg) = 60.2 km/h`

Part d)

average velocity must be ZERO

Step-by-step explanation:

While he drive San Antonio to Houston

Half the time move with speed 51 km/h and next half his speed is 71 km/h

so we have

`v_1(T/2) + v_2(T/2) = d`

`T = (2d)/(v_1 + v_2)`

`T = (2d)/(51 + 71)`

so average speed is given as

`v_(avg) = (d)/(T)`

`v_(avg) = (d)/((2d)/(51 + 71))`

`v_(avg) = 61 km/h`

Part b)

While he move from Houston to San Antonio half the distance is moved with 51 km/h and next half distance with 71 km/h

so we have

`T = (d)/(2v_1) + (d)/(2v_2)`

`T = (d)/(102) + (d)/(142)`

so average speed is given as

`v_(avg) = (d)/( (d)/(102) + (d)/(142))`

`v_(avg) = 59.4 km/h`

Part c)

Average speed for entire trip

`v_(avg) = (2d)/((2d)/(51 + 71) + (d)/(102) + (d)/(142))`

`v_(avg) = 60.2 km/h`

Part d)

Since total displacement of entire trip is zero

so average velocity must be ZERO