# Physics Question pls help!!!?

During a trip, a bus travels 11km with an average velocity of 21m/s, but then travels 1.0km at a smaller velocity of 4.2m/s, due to the presence of highway construction. Determine the average velocity of the bus for the entire trip.

its prob an easy question but i cant think right now cause i havent done pysics in a while pls help

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The first thing you'll want to find is how long the bus travelled at each velocity.

We know the bus travelled 11 km at 21 m/s, and that 11 km = 11,000 m. Therefore, travelling at 21 m/s:

Distance = speed x time

11,000 = 21 x time

11,000/21 = time

We could change 11,000/21 into a decimal answer (~523.8 sec), but it’s more accurate to leave it as 11,000/21.

Also, we know that the bus travelled 1.0 km at 4.2 m/s, and that 1.0 km = 1000 m. To find out how much time the bus travelled at 4.2 m/s:

Distance = speed x time

1000 = 4.2 x time

1000/4.2 = time

Therefore, the total time the bus spent travelling was 11,000/21 + 1000/4.2 = about 761.90 seconds.

About (11000 / 21) / 761.90 = proportion of time spent travelling at 21 m/s.

About (1000 / 4.2) / 761.90 = proportion of time spent travelling at 4.2 m/s.

To find the average velocity, just multiply each velocity with the corresponding proportion of time that the bus travelled at that velocity. Then, add the two products.

21 x (11000 / 21) / 761.90 = 14.4375

4.2 x (1000 / 4.2) / 761.90 = 1.3125

14.4375 + 1.3125 = about 15.75 m/s

If you look at the numbers you were given (11km, 21m/s, 1.0km, and 4.2 m/s), you'll notice that each of them has two significant figures. Therefore, we round 15.75 m/s to 2 sig figs.

The average velocity is 16 m/s.

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• (11,000/21) = 523.809s.

(1,000/4.2) = 238.095s.

Total = 761.9s.

Total distance = 12,000m.

(12,000m/761.9s) = 15.75m/sec.

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