Solve accelerated movement with time loss?
During a car race, a car's speed is 240 km/h. Due to a penalty, the driver with a constant acceleration of -30 m/s² reduces his speed to 80 km/h and drives at exactly this speed for 300 m. He then accelerates steadily and reaches his original speed after 4.0 s.
Determine the minimum time loss that this penalty means, assuming that the vehicle would have moved uniformly at a constant speed of 240 km/h without the penalty.
I'm currently practicing for an upcoming test, as mentioned in the other questions. Unfortunately, there's no online solution to this problem. I came up with a time loss of about 6 seconds. I calculated the time segments for each segment, then calculated the distance in each segment, and differentiated the different times for each segment with and without a penalty. I just wanted to ask if the result is correct.
thanks to those who calculate this 🙂 I am writing the paper tomorrow and an answer would be a great help
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v1 = 240 km/h = 2400/36 m/s
v2 = 80 km/h = 800/36 m/s
The braking process takes t = (v2-v1)/a = 40/27 ~ 1.48 sec.
At this time, the carriage s = 1/2 * a * t2 = 8000/243 ~ 32.92 meters.
For 300 m with v = 800/36 m/s, the carriage 27/2 = 13.5 sec
The carriage accelerates in 4 seconds from v1 to v2, a = (v2-v1)/4 = 100/9
In this time the carriage s = 1/2 * a * t2 = 800/9 ~ 88.88 meters is returned.
Total distance: 8000/243 + 300 + 800/9 = 102500/243 ~ 421.81 m
Total time: 40/27 + 27/2 + 4 = 1025/54 ~ 18.98 sec
Without delay, the car would have needed 1025/162 ~ 6.32 sec.
Makes a difference of 1025/54 – 1025/162 = 1025/81 ~ 12.65 sec
Section 1)
t_1 = 1.48 s; s_1 = 65.8 m
Section 2)
t_2 = 13.5 s; s_2 = 300 m
Section 3)
t_3 = 4 s; s_3 = 177.78 m
Total:
t_ges = 18.98 s; s_tot = 543.56 m
without delay: with v = (240 / 3.6) m/s2 and s = 543.56 m, the 8.15 s
Δt = 10.83 s
I have a question: how to get to the 65m in the first section? I have calculated S=a/2*t^2+Vo*t and used for Vo 240km/h, since a speed difference is created and that was the initial speed. Is that wrong?
Okay, I'm really tired. I found the error, had positively written the value a when braking. Had little braking and didn't know you had to leave it negative. Thanks for the help :))
t_1 = ((80 / 3.6) m/s – (240 / 3.6) m/s) / -30 m/s2 = 1.48 s
s_1 = (((((80 / 3.6) m/s + (240 / 3.6) m/s) / 2) * 1.48 s = 65.8 m
Thank you 🙂
Great success at work!
I'm coming to 10.8 s.
For the 300 m alone, it needs 13.5 s instead of 4.5 s, which means it already lacks at least 9 s!
Yeah, I did. I only added the 13.5s with 1.48 and 4s. Without the penalty, he would have driven the total 866m in 13s approx. The difference is then about 6 seconds.
However, it travels 'only' 300 m (plus distance to braking + distance to acceleration). For the 300 m it needs 13.5 s. If he would drive the 300 m at 240 km/h, he would only need 4.5 s! He's losing 9 s alone! The phases for braking and acceleration come to this!
Please get used to decent writing, eg with units.
The routes from the 2nd and 3rd sections, which are not from the 1st! It brakes in 1.48 s and resets 388 m while it needs 178 m for 4 s acceleration?
And even if: 478 m at 240 km/h are 7.17 s, but it already needs 17.5 s for sections 2 and 3! Are already 10.33 s time lost!
But if I'm honest, I don't really see the mistake. I think I might have had a mistake in thinking in general
I have reckoned again: in the case of an incision 1 (without penalty) come to 5.8s, on the second as I said to 4.5s and on the third to 2.67s. I assume that he'll take the same path as the road with punishment. Therefore I also came to the total 866m: 388+300+178
I think I see my mistake. I'll go back. Thank you.