16 Jun 2019
Leaving Cert Applied Maths Higher Level 1980-1992 Solutions By Maths Grinds
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Leaving Cert Applied Maths Higher Level 1980 Solution
- (i) $0\vec{i}+\left(14-8\sqrt{2}\right)\vec{j}$, $\frac{7+4\sqrt{2}}{8}}$ hours (ii) 1.45 hours
- $\frac{1}{20} – \frac{T_1}{10W}$
- (ii) $\frac{\sqrt{5}u^2}{4g}$
- (ii) $\frac{W \sin\left(\alpha+\lambda\right)}{\cos \lambda\right)}$ (iii) $W \tan \left(\alpha+\lambda\right)}$ (iv) $W\sin\left(\alpha+\lambda\right)$
- (b) $ \mu \sin \left( \alpha \right) \left( 1+e \right) $
- $ x = \frac{2f}{k} $
- $\frac{3l \pm \sqrt{5}l}{6}$
- $1.7$ m, $\frac{\pi}{2}$ s, $0.1477$ s
- (a) $y=\sqrt{2\sqrt{2} x – 3}$ (b) $2.18$ m s$^{-1}$
- (a) $\frac{12}{7}$ (b) $\frac{4}{5}$
Leaving Cert Applied Maths Higher Level 1981 Solution
- (i) $1.2$ m s$^{-2}$ and $-2$ m s$^{-2}$
- “Show that” question
- “Show that” question
- “Show that” question
- $\frac{7}{9} W$, $W$ and $\frac{5}{9}W$
- (a) $12.1$ percent shorter (b) $7$ rad s$^{-1}$
- $\omega = \sqrt{\frac{20g}{19 r}}$
- (b) 66
- (b)(ii) $T=\frac{W}{4}$ and $B=\frac{3}{4}W$ (ii) $\frac{2}{3}$
- (a) $y=\frac{24}{24 \sin x – 8 \sin^3 x + 1}$ (b) $28.76$ m
Leaving Cert Applied Maths Higher Level 1982 Solution
- (i) (a) $80$ m (ii) $150$ s (b) $0.5094$ N
- (i) $25\vec{i} – 15\sqrt{3}\vec{j}$ (ii) $1.92$ km
- (ii) $\frac{\sqrt{3}}{2}$ (iii) $\frac{400}{7g}$ m
- (a) $\frac{5}{4}$ and $\frac{35}{4}$ m s$^{-1}$, $187.5$ kg m s$^{-1}$ (b) $\frac{1}{3}$
- “Show that” question
- $\frac{W}{\sqrt{3}}$ for both, $\mu \geq \frac{1}{\sqrt{3}} $
- (ii) $11 ml^2 $
- $\omega = \frac{5}{2}$ rad s$^{-1}$ and $0.257 $ s
- (a) (i) $500g$ Pa (ii) $\frac{125\pi g}{16}$ N (iii) 7:1 (b) $0.9$
- (a) $y=\left(\frac{1+x^3}{4}\right)^{\frac{1}{3}}$ (b) $s=\ln \left(1+t\right)$ and $s=\ln 2$ m
