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a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a displacement of 215 km, east. what is the driver’s average velocity?

asked
User Demetri
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1 Answer

3 votes

Answer:

v = 98.75 km/h

Step-by-step explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

v₁ = d₁/t₁

= 135/1.5

= 90 km/h

The average speed of vehicle after stopping

v₂ = d₂/t₂

= 215/2

= 107.5 km/h

The total average velocity of the driver

v = (v₁ +v₂) /2

= (90 + 107.5)/2

= 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

answered
User Zakwan
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9.4k points