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# While playing snowbarding a guy is going down a hill with 120 kmh. He then ends at the flat part of the hill, and his speed decreases with 6

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While playing snowbarding a guy is going down a hill with 120 kmh. He then ends at the flat part of the hill, and his speed decreases with 6.5m/s^2 . Calculate the minumum lenght of the flat part.

physics
Guest Jan 5, 2015

#1
+26965
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Use v2 = u2 + 2*a*s  where v = final speed (0 m/s), u = initial speed (120*103/3600 m/s) a = acceleration ( -6.5 m/s2) and s = distance (in metres).

Rearrange as: s = (v2 - u2)/(2a)

$${\mathtt{s}} = {\frac{\left({\mathtt{0}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{120\,000}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6.5}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{85.470\: \!085\: \!470\: \!085\: \!470\: \!1}}$$

or s ≈ 85.5 m

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Alan  Jan 6, 2015
#1
+26965
+5

Use v2 = u2 + 2*a*s  where v = final speed (0 m/s), u = initial speed (120*103/3600 m/s) a = acceleration ( -6.5 m/s2) and s = distance (in metres).

Rearrange as: s = (v2 - u2)/(2a)

$${\mathtt{s}} = {\frac{\left({\mathtt{0}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{120\,000}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6.5}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{85.470\: \!085\: \!470\: \!085\: \!470\: \!1}}$$

or s ≈ 85.5 m

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Alan  Jan 6, 2015