Looking for d=rt or something else?

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The discussion revolves around calculating the distance traveled by a car accelerating from 20 m/s to 40 m/s over 10 seconds. The correct approach involves using the average speed, which is determined to be 30 m/s, leading to a distance of 300 meters when multiplied by the time. The initial confusion stemmed from miscalculating the average speed and the arithmetic involved. Both methods discussed—using average speed and the kinematic equation—yield the same result of 300 meters. Ensuring consistent units and accurate calculations is emphasized for solving similar physics problems.
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I am having so much trouble. The kind folks here have already helped me bunches with free-fall problems. Now I think I have a new problem.

Q. A 100-kg auto enters a freeway onramp at 20 m/s and accelerates uniformly up to 40 m/s in a time of 10 s. How far does the auto travel during that time?

I am unsure what I am looking for here...I know I am seeking distance, but am I seeking it through d=rt, where I discern the rate by going 40m/s - 20 m/s divided by 2 to get average rate (10 m/s), and then plugging that into d=10m/s x 10 s? I get 100, but...That looks wrong to me, my units look wrong...but then, I try d=1/2at**2, so that d=1/2 ? (10)**2, where I am just guessing at acceleration by saying 40m/s-20m/s to get 20, giving me 1000, that's crazy...I am lost, hopelessly lost. Or is this where I have to get v=at before I can proceed onto another equation?
 
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d= rt is only true as long as "r" (rate of change, speed) is a constant. If acceleration is a constant, then δr= at and d= (1/2)at2+ v0t. If the vehicle accelerates from 20 m/s to 40m/s then δr= 40-20= 20. Since that happens in 10s.,a= 20/10= 2m/s2. Now putting that into d= (1/2)at2+v0t gives d= (1/2)(2)(10)2+ 20(10)= 100+ 200= 300 m.

Actually, the way you were doing it is just fine. As long as the acceleration is constant, you can use "average speed". The initial speed was 20 m/s and final speed was 40 m/s so the "average speed" is (40+20)/2= 30 m/s. That average speed for 10 seconds gives 30(10)= 300 m as above.

It's not your physics that needs work, it's your arithmetic! The "average" of 20 and 40 is (40+20)/2= 30 not "(40-20)/2"!
 


Hi there,

It seems like you are on the right track! In this problem, you are looking for the distance traveled by the car during the given time period. So, you are correct in using the equation d=rt, where r is the average rate of acceleration (which you correctly calculated as 10m/s) and t is the time period of 10 seconds. When you plug in these values, you get 100m as the distance traveled by the car.

Alternatively, you can also use the equation d=1/2at^2, where a is the acceleration and t is the time period. In this case, you correctly calculated the acceleration as 20m/s^2 (by dividing the change in velocity by the time period), which gives you a distance of 100m as well.

So, both methods give you the same answer of 100m. It's always a good idea to double check your units to make sure they are consistent. In this case, you have the correct units of meters for distance, seconds for time, and meters per second for acceleration.

I hope this helps and good luck with your future physics problems!
 
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