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.

Guest Jan 5, 2015

#1**+5 **

Use v^{2} = u^{2} + 2*a*s where v = final speed (0 m/s), u = initial speed (120*10^{3}/3600 m/s) a = acceleration ( -6.5 m/s^{2}) and s = distance (in metres).

Rearrange as: s = (v^{2} - u^{2})/(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

.

Alan
Jan 6, 2015

#1**+5 **

Best Answer

Use v^{2} = u^{2} + 2*a*s where v = final speed (0 m/s), u = initial speed (120*10^{3}/3600 m/s) a = acceleration ( -6.5 m/s^{2}) and s = distance (in metres).

Rearrange as: s = (v^{2} - u^{2})/(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

.

Alan
Jan 6, 2015