in an auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side on the track at 71.5m/s. The driver of the Thunderbird realizes that she has to make a pit stop, so she smoothly slows to a stop over a distance of 250m. After spending 5.0s in the stop, she then accelerates back to her original 71.5m/s over a distance of 350m. At this point, how far behind the mercedes is she?
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Accessories March 18th 2011
March 18th, 2011 at 2:18 am
i’ll guess and say 958 meters behind.
March 18th, 2011 at 2:52 am
The easiest way to answer this question.
If the Thunderbird has to stop and the velocity goes from 71.5 m/s to 0 m/s “smoothly” then the average velocity is 71.5 m/s / 2 or 35.75 m/s. If it takes her 250 m to stop with an average velocity of 35.75 m/s the time is d/v or 250 m / 35.75 m/s = 7.0 s
She stops for 5.0 s
She then speeds back up. Again the average velocity is 35.75 m/s and this time the distance is 350 m, so the time is d/v or 350 m / 35.75 m/s = 9.8 s
So, the pit stop take a total of 7.0 s + 5.0 s + 9.8 s = 21.8 s
The Mercedes kept a constant 71.5 m/s during that 21.8 s so it could be as far as 1,560 m ahead of the Thunderbird. The problem is a bit ambiguous, though. It does not state if the 250 m of stopping and the 350 m of starting were in the same direction that the Mercedes was moving. The answer of 1.56 km assume that the 250 m and 350 m were not in the same direction. If those distances are in the same direction, the we would have to decrease the 1,560m - 250m -350m = 960 m.
So, the Thunderbird is somewhere between 960 m and 1560 m behind. If you instructor says one of the answers is correct, then they are wrong. You cannot give a specific answer with the problem the way it is worded.
March 18th, 2011 at 3:51 am
well, you will need to find the total time the thunderbird was out of the race using the equations of motion, then u can calculate how far the mercedes has travelled during that time
so…
distance = 1/2 (V initial + V final) x time
250 = 1/2 (71.5) t
250/35.75 = t
6.99s = time it took the thunderbird to reach the stop
add 5 seconds to it because “she spent 5s in the stop”
now find the time it took her to accelerate into the race
distance = 1/2 (V initial + V final) x time
350 = 1/2 (71.5) x t
350/35.75 = t
9.29s = t
total time = 6.99 + 5 + 9. 29
sub your total time in for the mercedes (btw im assuming mercedes is moving at a constant speed)
distance = velocity x time
= 71.5 m/s x 21
= 1557.5 meters
or at least thats how i think its done….