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Question Number 216621 by Tawa11 last updated on 12/Feb/25

Answered by A5T last updated on 12/Feb/25

Commented by A5T last updated on 12/Feb/25

[OA_1 DT]=[ABCT] when both cars meet at time t ⇒20t=((30(t−20+t−10))/2)⇒4t=3(2t−30) ⇒t=45s⇒ s=20×45=900m ⇒Sports car overtakes first car 45s and 900m after car overtakes stationary sports car.

$$\left[\mathrm{OA}_{\mathrm{1}} \mathrm{DT}\right]=\left[\mathrm{ABCT}\right]\:\mathrm{when}\:\mathrm{both}\:\mathrm{cars}\:\mathrm{meet}\:\mathrm{at}\:\mathrm{time}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{20t}=\frac{\mathrm{30}\left(\mathrm{t}−\mathrm{20}+\mathrm{t}−\mathrm{10}\right)}{\mathrm{2}}\Rightarrow\mathrm{4t}=\mathrm{3}\left(\mathrm{2t}−\mathrm{30}\right) \\ $$$$\Rightarrow\mathrm{t}=\mathrm{45s}\Rightarrow\:\mathrm{s}=\mathrm{20}×\mathrm{45}=\mathrm{900m} \\ $$$$\Rightarrow\mathrm{Sports}\:\mathrm{car}\:\mathrm{overtakes}\:\mathrm{first}\:\mathrm{car}\:\mathrm{45s}\:\mathrm{and}\:\mathrm{900m}\: \\ $$$$\mathrm{after}\:\mathrm{car}\:\mathrm{overtakes}\:\mathrm{stationary}\:\mathrm{sports}\:\mathrm{car}. \\ $$

Commented by Tawa11 last updated on 12/Feb/25

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 12/Feb/25

Commented by mr W last updated on 12/Feb/25

distance is the area under the corresponding v−t curve. d=20x=((30×10)/2)+30(x−20) ⇒x=45 s ⇒d=20×45=900 m

$${distance}\:{is}\:{the}\:{area}\:{under}\:{the} \\ $$$${corresponding}\:{v}−{t}\:{curve}. \\ $$$${d}=\mathrm{20}{x}=\frac{\mathrm{30}×\mathrm{10}}{\mathrm{2}}+\mathrm{30}\left({x}−\mathrm{20}\right) \\ $$$$\Rightarrow{x}=\mathrm{45}\:{s} \\ $$$$\Rightarrow{d}=\mathrm{20}×\mathrm{45}=\mathrm{900}\:{m} \\ $$

Commented by Tawa11 last updated on 12/Feb/25

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 14/Feb/25

Sirs, can this be solved without graph?

$$\mathrm{Sirs},\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{without}\:\mathrm{graph}? \\ $$

Commented by mr W last updated on 15/Feb/25

the solution above is without graph! or you don′t understand it without the graph?

$${the}\:{solution}\:{above}\:{is}\:{without} \\ $$$${graph}!\:{or}\:{you}\:{don}'{t}\:{understand}\:{it} \\ $$$${without}\:{the}\:{graph}? \\ $$

Commented by Tawa11 last updated on 15/Feb/25

I mean, can we solve analytically without solving from graph? Can we solve the distance and time without drawing graph (v−t graph)?

$$\mathrm{I}\:\mathrm{mean},\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{analytically} \\ $$$$\mathrm{without}\:\mathrm{solving}\:\mathrm{from}\:\mathrm{graph}? \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{and}\:\mathrm{time} \\ $$$$\mathrm{without}\:\mathrm{drawing}\:\mathrm{graph}\:\left(\mathrm{v}−\mathrm{t}\:\mathrm{graph}\right)? \\ $$

Commented by mr W last updated on 15/Feb/25

they are exactly given above. d_1 =20x d_2 =((30×10)/2)+30(x−20) with x=time

$${they}\:{are}\:{exactly}\:{given}\:{above}. \\ $$$${d}_{\mathrm{1}} =\mathrm{20}{x} \\ $$$${d}_{\mathrm{2}} =\frac{\mathrm{30}×\mathrm{10}}{\mathrm{2}}+\mathrm{30}\left({x}−\mathrm{20}\right) \\ $$$${with}\:{x}={time} \\ $$

Commented by Tawa11 last updated on 15/Feb/25

Ohh, I grab sir. Thanks sir.

$$\mathrm{Ohh},\:\mathrm{I}\:\mathrm{grab}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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