How Fast Must the Car Travel in the Second Half to Average 51 km/h?

AI Thread Summary
To average 51 km/h over a 94 km journey, the car must complete the first 47 km at 39 km/h, taking approximately 1 hour. The total time for the trip at the desired average speed is 1.52 hours, leaving 0.52 hours for the second half. By calculating the time taken for the first half and subtracting it from the total time, the required speed for the second half can be determined. The calculations suggest that the speed needed for the second half is around 90.4 km/h, which some participants believe may be too high. The discussion emphasizes the importance of using total distance and average velocity to find the correct speed for the second half of the trip.
mike5754
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Homework Statement



A car making a 94-km journey travels 39 km/h for the first 47 km. How fast must it go during the second 47 km to average 51 km/h?

Homework Equations





The Attempt at a Solution

 
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don't want someone to solve it for me. Just point me in the right direction and help out a little bit.
 
Torquescrew said:
Hey, dude. Try using the following formula:
Average Velocity = (Final Velocity + Initial Velocity)/2

Why /2?
I have found that to go 94 km at an average speed of 51 km/h that would be 14.17m/s.
Thus if you went 47 km at 39 km/h you would be moving at 10.83 m/s.
So with this data averaging 51 km/h the entire trip would take 1.52 hours. And the first 47 km (half) would have taken 1 hour. Am I correct so far? So to find the change in time we are looking for it would be 1.52-1.00 giving you .52 hours.

I have Average velocity defined as deltaX/deltaT

deltaX = 47km and delta T = .52 Hours

Thus giving me 90.4 km/h

does this sound right? seems a bit high but maybe I am wrong, this is my first physics class ever. Thanks
 
Torquescrew said:
Hey, dude. Try using the following formula:
Average Velocity = (Final Velocity + Initial Velocity)/2

Don't think that's necessarily right. Average velocity is total distance / total time...
 
mike5754 said:
Why /2?
I have found that to go 94 km at an average speed of 51 km/h that would be 14.17m/s.
Thus if you went 47 km at 39 km/h you would be moving at 10.83 m/s.
So with this data averaging 51 km/h the entire trip would take 1.52 hours. And the first 47 km (half) would have taken 1 hour. Am I correct so far? So to find the change in time we are looking for it would be 1.52-1.00 giving you .52 hours.

I have Average velocity defined as deltaX/deltaT

deltaX = 47km and delta T = .52 Hours

Thus giving me 90.4 km/h

does this sound right? seems a bit high but maybe I am wrong, this is my first physics class ever. Thanks

I'm not tracking what you are doing. Find t1 = the time to the midway point. Calculate the total time t1+t2 from the total distance and total average velocity. That gives you t2, and you know the distance for the 2nd half of the trip...

(and no, the answer is not what Torquescrew suggested)
 

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