Motion In A Straight Line Practice Questions

${\text{t}}_{ {\text{AB}}}=\;{ {\text{x}}}/{5 {\text{m}} / {\ s}}$In motion ${\text{BC}}$ ${\text{x}}= {\text{d}}_1+ {\text{d}}_2$ where {\text{d}}_1 & {\text{d}}_2 we the distance travelled with 10 ${\text{m}} / {\ s}$ and $15 {\text{m}} / {\ s}$ respectively in equal time intervals $\;\frac{\;\text{ 't' }\;}{2}$ each$\; {\text{d}}_1=\;{10 {\text{t}}}/{2}, {\text{d}}_2=\;{15 {\text{t}}}/{2}$ $\; {\text{d}}_1+ {\text{d}}_2= {\text{x}}=\;{ {\text{t}}}/{2}(10+15)=\;{25 {\text{t}}}/{2}$ $\;< {\text{v}}>=\;{2 {\text{x}}}/{{ {\text{x}}}/{5}+\;{2 {\text{x}}}/{25}}=\;\frac{2 \times 25}{5+2}=\;\frac{50}{7} {\text{m}} / {\ s}$Ans. : 50

$\; {\text{v}}_{ {\text{i}}}=\sqrt{2 {\text{gh}}_{ {\text{i}}}}$ $\;={\sqrt{2 \times 10 \times 9.8}} ↓$ $\;=14 {\text{m}} / {\ s} ↓$ $\; {\text{v}}_{ {\text{f}}}=\sqrt{2 {\text{gh}}_{ {\text{f}}}}$ $\;={\sqrt{2 \times 10 \times 5}} ↑$ $\;=10 {\text{m}} / {\ s} ↑$ \; |{ \vec{\text{a}}_{\;\text{avg }\;}|= |\;{\Delta { \vec{\text{v}}}/{\Delta {\text{t}}}|=\;\frac{24}{0.2}=120 {\text{m}} / {\ s}^2

(A) and (E) onlySlope of $A=V_A$Slope of $B=V_B$(slope $)_B > (\;\text{ slope }\;)_A$ ${\text{V}}_{ {\text{B}}} > {\text{V}}_{ {\text{A}}}$ $\therefore {\text{t}}_{ {\text{B}}} < {\text{t}}_{ {\text{A}}}$

Using ${\text{I}}^{ {\ st}}$ equation of motion$\;t=\;\frac{v-u}{a}$ $\;t=\;\frac{60-10}{2}=\;\frac{50}{2}=25 {\ sec}$

Time taken b / w A & B \Rightarrow t_1=\;\frac{x}{v_1}Time between b / w \;\; B & C \Rightarrow t_2=\;{ {\text{x}}}/{ {\text{v}}_2}Average velocity $(v)=\;\frac{\;\text{ Total displacement }\;}{\;\text{ Total time }\;}=\;\frac{x+x}{t_1+t_2}=\;\frac{2 x}{\frac{x}{v_1}+\;\frac{x}{v_2}}$(v) $=\;{2 {\text{v}}_1 {\text{v}}_2}/{ {\text{v}}_1+ {\text{v}}_2} \;\;$ or $\;\; \;{2}/{ {\text{v}}}=\;{1}/{ {\text{v}}_1}+\;{1}/{ {\text{v}}_2}$

${\text{S}}=2.5 {\text{t}}^2$At, ${\text{t}}=5 {\sec}$.${\text{v}}=5 \times 5=25 {\text{ms}}^{-1}$

$\;\;\text{ Average velocity }\;=\;\frac{\;\text{ Total displacement }\;}{\;\text{ Total time }\;}$ $\;=\;{ {\text{x}}+ {\text{x}}}/{{ {\text{x}}}/{ {\text{v}}_1}+\;{ {\text{x}}}/{ {\text{v}}_2}}=\;{2 {\text{v}}_1 {\text{v}}_2}/{ {\text{v}}_1+ {\text{v}}_2}$

$\;\;\text{ Given, }\; \;\; x= (5 t^{2-}4 t+5) m$ $\;V=\;\frac{d x}{d t}=10 t-4$ $\;t=2, V=20-4$ $\;V=16 m / s$

$\;\;{2}/{ {\text{V}}_{ {\text{av}}}}=\;\frac{1}{3}+\;\frac{1}{5}=\;\frac{8}{15}$ $\;\Rightarrow \;\; {\text{V}}_{ {\text{av}}}=\;\frac{15}{4}=3.75 {\text{km}} / {\text{h}}$

$\;{ {\text{dx}}}/{ {\text{dt}}}=$ slope $\ge 0$ always increasing(A - II)$\;{ {\text{dx}}}/{ {\text{dt}}} < 0$; and at ${\text{t}} \to \infty \;{ {\text{dx}}}/{ {\text{dt}}} \to 0$(B - IV)$\;{ {\text{dx}}}/{ {\text{dt}}} > 0$ for first half $\;{ {\text{dx}}}/{ {\text{dt}}} < 0$ for second half.(C - III)$\;{ {\text{dx}}}/{ {\text{dt}}}=$ constant(D - I)

${\text{AB}}=x$ ${\text{BC}}=x$ $2 x+ {\text{CD}}=3 x$ $\; {\text{CD}}=x$ $\;⟨ {\text{v}}⟩=\;{3 x}/{\frac{x}{v_{1}}+\;\frac{x}{v_{2}}+\;\frac{x}{v_{3}}}=\;\frac{3 v_{1} v_{2} v_{3}}{v_{2} v_{3}+v_{1} v_{3}+v_{1} v_{2}}$

${ {\text{v}}}^{\to }=\;{ {\text{d}} { {\ s}}^{\to }}/{ {\text{dt}}} \Rightarrow ∫ {\text{d}} { {\ s}}^{\to }=∫ { {\text{v}}}^{\to } {\text{dt}}$Area of $\vec{v}$ vs time gives displacement$\vec{a}=\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}} \Rightarrow ∫ {\text{d}} { {\text{v}}}^{\to }=∫ { {\text{a}}}^{\to } {\text{dt}}$Area of $\vec{a}$ vs t graph gives change in velocity

$\; {\text{V}}_{ {\text{A}}}=108 \times \;\frac{5}{18}=30 {\text{m}} / {\ s}$ $\; {\text{V}}_{ {\text{B}}}=72 \times \;\frac{5}{18}=20 {\text{m}} / {\ s}$ $\; {\text{T}}_{ {\text{A}}}=\;{ℓ+ {\text{L}}}/{30}, {\text{T}}_{ {\text{B}}}=\;{4 ℓ+ {\text{L}}}/{20}$ $\; {\text{T}}_{ {\text{A}}}= {\text{T}}_{ {\text{B}}}-35$ $\;\;{ℓ+ {\text{L}}}/{30}=\;{4 ℓ+ {\text{L}}}/{20}-35$ $\; {\text{Given}}, \;\; {\text{L}}=60 ℓ$ $\;\;\frac{61 ℓ}{30}=\;\frac{64 ℓ}{20}-35$ $\;\;\frac{192 ℓ-122 ℓ}{60}=35$ $\;70 ℓ=60 \times 35$ $\;ℓ=30 {\text{m}}$ $\; {\text{L}}=60 ℓ=1800 {\text{m}}$

$\; {\text{V}}_{ {\text{A}}}=\;{90 {\text{km}}}/{ {\text{hr}}}=25 {\text{ms}}^{-1}$ $\; {\text{V}}_{ {\text{B}}}=\;{54 {\text{km}}}/{ {\text{hr}}}=15 {\text{ms}}^{-1}$ $\;{ {\text{V}}_{ {\text{BA}}}^{\to }}={ {\text{V}}_{ {\text{B}}}^{\to }}-{ {\text{VA}}}^{\to }=40 {\text{ms}}^{-1}$ $\;\;\text{ Time of crossing }\;=\;\frac{\;\text{ Length of train }\;}{\;\text{ Re lative velocity }\;}$ $\;8=\;\frac{ℓ}{40}$ $\;ℓ=8 \times 40=320 \;\text{ meter }\;$

${\text{u}}=20 {\text{m}} / {\ s}, {\text{S}}_1=500 {\text{m}}, {\text{v}}=0$By third equation of mation$\;0=(20)^{2-}2 {\text{a}} .500 \Rightarrow {\text{a}}=\;\frac{4}{10} {\text{m}} / {\ s}^2$ $\; {\text{u}}=20 {\text{m}} / {\ s}, {\text{S}}_2=250 {\text{m}}, {\text{v}}=?$ $\; {\text{v}}^2=(20)^{2-}2 {\text{a}} .250$ $\;= {\text{v}}={\sqrt{200}} {\text{m}} / {\ s}$ $\; {\text{x}}=200$

${\text{u}}=0, \;\;$ Say acceleration is aFort s $\;\; 10=\;\frac{1}{2} {\text{at}}^2$For $2 {\text{t}} {\ s}$ $10+x=\;\frac{1}{2} a(2 t)^2$ $\;\frac{10+x}{10}=\;\frac{4}{1}$ $x=30 m$

Let they meet at ${\text{t}}= {\text{t}}$So first ball gets $t$ sec.$\& 2^{\;\text{nd }\;}$ gets $( {\text{t}}-2)$ sec. & they will meet at sameheight${\text{h}}_1=50 {\text{t}}-\;\frac{1}{2} {\text{gt}}^2$ ${\text{h}}_2=50( {\text{t}}-2)-\;\frac{1}{2} {\text{g}}( {\text{t}}-2)^2$ ${\text{h}}_1= {\text{h}}_2$ $50 {\text{t}}-\;\frac{1}{2} {\text{gt}}^2=50( {\text{t}}-2)-\;\frac{1}{2} {\text{g}}( {\text{t}}-2)^2$ $100=\;\frac{1}{2} g[t^{2-}(t-2)^2]$ $100=\;\frac{10}{2}[4 t-4]$ $5= {\text{t}}-1$ $t=6 {\sec} \;\text{. }\;$

$F_{R} d=\;\frac{1}{2} m v^{2}$ $\;\frac{d_{2}}{d_{1}}=(\;\frac{v_{2}}{v_{1}})^{2}=(\;\frac{1}{3})^{2}$ $d_{2}=d_{1} \times \;\frac{1}{9}=3 m$