Gravity and motion — Question 245

ANSWER

(a) A $6$-square east vector followed by an $8$-square north vector; resultant from O to $(6,8)$. (b) $10.0\,\mathrm m$, $53.1^\circ$ north of east.

Worked solution

(a) The second arrow starts at the first arrowhead. The resultant joins the start of the first vector to the end of the second, with its arrowhead at $(6,8)$.

(b) $R=\sqrt{6.00^2+8.00^2}=10.0\,\mathrm m$; $\theta=\tan^{-1}(8/6)=53.1^\circ$.

Equivalent physically justified methods, alternative valid sketches and consistent units accepted.

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Draw correctly oriented and scaled head-to-tail vectors.

[1 mark]

Part a: Draw the resultant from O to the final endpoint, with an arrowhead.

[1 mark]

Part b: Obtain the magnitude using Pythagoras.

[1 mark]

Part b: Obtain the angle and identify north of east.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.