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In 10 seconds car can accelerate to 100km/h. Let's say acceleration is constant, calculate the distance it will take the car to reach 90 km/h after leaving a built-up area where it was moving 60km/h.

$60\frac {km}{h}=16\frac{4}{6}\frac{m}{s}$
$90\frac {km}{h}=25\frac{m}{s}$
I calculated acceleration using $a=\frac {Vp-Vk}{t}=\frac {27\frac{7}{9}-0}{10}=2\frac{7}{9}$
$t=\frac{Vk-Vp}{a}=3$
and distance using $s=\frac {V1+V2}{2}\cdot(t2-t1)=\frac {16\frac{4}{6}+25}{2}\cdot3=[62\frac{1}{2}m]$

I think I got everything right but the result is slightly off, it should be [61,8m] (according to notes)
$s=Vp\cdot t+\frac{a\cdot t^2}{2}$ gives me the same result
What am I doing wrong?

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    $\begingroup$ Perhaps the conversion from km/h to m/s wasn't the same as was given by the answer. $\endgroup$ – AfronPie Jan 12 at 22:05
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    $\begingroup$ When working in the distance domain instead of the time domain, using conservation of energy often makes the calculations simpler. $\endgroup$ – amd Jan 12 at 22:22
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    $\begingroup$ The exact answer is $125/2$m, as you computed. Truncating values early in the computation instead of carrying exact values all the way through could account for the discrepancy. If the notes don’t include details of the calculation used to produce 61.8, I wouldn’t spend much time trying to reverse-engineer it. $\endgroup$ – amd Jan 12 at 22:34
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    $\begingroup$ @b-czostek You calculated the acceleration, time and the distance correctly. The distance result 61.8 m from notes is incorrect. $\endgroup$ – yW0K5o Jan 17 at 18:55

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