Vraag 10

uitkomst:$s=4{,}8 \mathrm{~m}(met een marge van$0{,}2 \mathrm{~m})

voorbeelden van een bepaling:

methode 1

De afstand waarover de springer valt, is gelijk aan de oppervlakte onder de$(v, t)-grafiek tot$t=1{,}0 \mathrm{~s}. De afstand van de val is dan gelijk aan$s=\frac{1}{2} \cdot 1{,}0 \cdot 9{,}6=4{,}8 \mathrm{~m}.

of

methode 2

De afstand waarover de springer valt, is uit te rekenen met s=vts=vs=smet v=v_{gem}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8\mathrm{~}ms^{-1}v=v_{gem}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8\mathrm{~c}ms^{-1}v=v_{gem}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8\mathrm{~cm}ms^{-1}v=v_{gem}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}81\mathrm{~cm}ms^{-1}v=v_{gem}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{gemM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{gemEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{gemGEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{geGEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{gGEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-1}v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{-}v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8ms^{\placeholder{}}v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8msv=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8mv=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}8v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4{,}v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=4v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)=v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{2}\right)v=v_{GEM}=\left(\frac{\left(9{,}6+0\right)}{\placeholder{}}\right)v=v_{GEM}=\left(\left(9{,}6+0\right)\right)v=v_{GEM}=\left(\left(9{,}6+0\right)\right)v=v_{GEM}=\left(\left(9{,}6+\right)\right)v=v_{GEM}=\left(\left(9{,}6\right)\right)v=v_{GEM}=\left(\left(9{,}\right)\right)v=v_{GEM}=\left(\left(9\right)\right)v=v_{GEM}=\left(\left(\right)\right)v=v_{GEM}=\left(\right)v=v_{GEM}=v=v_{GEM}v=vgem_{GEM}v=vgem_{GEM}v=vgem_{GEMR}v=vgem_{GER}v=vgem_{GR}v=vgem_{GR}v=vgem_{\placeholder{}}v=vgemv=vgev=vgv=vv=vt. De vrije val duurt t_{v}=1{,}0\mathrm{~}st_{v}=1{,}0st_{v}=1{,}0t_{v}=1{,}t_{v}=1t_{v}=t_{v}t_{\placeholder{}}t, dus: s=vt=4{,}8\cdot1{,}0=4{,}8\mathrm{~m}

Opmerking Als gerekend is met$s=v twaarin voor$vniet de gemiddelde snelheid is ingevuld: geen scorepunten toekennen.