Question

Starting from home, you bicycle 25 km north in 2.4 h and then turn around and pedal straight home in 1.4 h . What are your (a) displacement at the end of the first 2.4 h , (b) average velocity over the first 2.4 h , (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip

Answer 1

Answer:

Part a)

$d = 25 km$

Part b)

$v_{avg} = 10.42 km/h$

Part c)

$v_{avg} = 17.86 km/h$

Part d)

Displacement for entire trip = 0

Part e)

Average velocity for entire trip will be zero

Explanation:

Part a)

Displacement after t = 2.4 hours is the straight line distance between initial and final positions

so we have

$d = 25 km$

Part b)

Average velocity is defined as

$v_{avg} = \frac{displacement}{time}$

$v_{avg} = \frac{25 km}{2.4 h}$

$v_{avg} = 10.42 km/h$

Part c)

During his return journey the displacement will be same

$displacement = 25 km$

$time = 1.4 h$

so average velocity is defined as

$v_{avg} = \frac{25 km}{1.4 h}$

$v_{avg} = 17.86 km/h$

Part d)

Displacement for entire trip = 0

as initial and final position will be same

Part e)

Average velocity for entire trip will be zero