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Question Number 69827 by Learner-123 last updated on 28/Sep/19
The acceleration of a particle moving in a straight line is defined as a=6t−20 m/s^2 , where t is in seconds. Knowing that s=0m when t=3s and that t=5sec when v=2m/s. Determine the total distance travelled when t=11s.
$${The}\:{acceleration}\:{of}\:{a}\:{particle}\:{moving} \\ $$$${in}\:{a}\:{straight}\:{line}\:{is}\:{defined}\:{as}\:{a}=\mathrm{6}{t}−\mathrm{20} \\ $$$${m}/{s}^{\mathrm{2}} ,\:{where}\:{t}\:{is}\:{in}\:{seconds}.\:{Knowing} \\ $$$${that}\:{s}=\mathrm{0}{m}\:{when}\:{t}=\mathrm{3}{s}\:{and}\:{that}\:{t}=\mathrm{5}{sec} \\ $$$${when}\:{v}=\mathrm{2}{m}/{s}.\:{Determine}\:{the}\:{total} \\ $$$${distance}\:{travelled}\:{when}\:{t}=\mathrm{11}{s}. \\ $$
Answered by Rio Michael last updated on 28/Sep/19
v = ∫adt = ∫(6t −20)dt = 3t^2 −20t + c v = 3t^2 −20t + c but t= 5, v =2 2 = 5 − 100 + c ⇒ c = 2 − 5 + 100 c = 97 ⇒v = 3t^2 −20t + 97 s = ∫vdt = ∫(3t^2 −20t + 97)dt = t^3 −10t^2 + 97t + c s = 0 at t= 3 solve and substitude for t = 11s
$$\:{v}\:=\:\int{adt} \\ $$$$\:\:\:\:=\:\int\left(\mathrm{6}{t}\:−\mathrm{20}\right){dt}\:=\:\mathrm{3}{t}^{\mathrm{2}} \:−\mathrm{20}{t}\:+\:{c} \\ $$$${v}\:=\:\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}\:+\:{c}\:{but}\:\:{t}=\:\mathrm{5},\:{v}\:=\mathrm{2} \\ $$$$\mathrm{2}\:=\:\mathrm{5}\:−\:\mathrm{100}\:+\:{c} \\ $$$$\Rightarrow\:{c}\:=\:\mathrm{2}\:−\:\mathrm{5}\:+\:\mathrm{100} \\ $$$$\:\:\:\:\:{c}\:=\:\mathrm{97} \\ $$$$\Rightarrow{v}\:=\:\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}\:+\:\mathrm{97} \\ $$$${s}\:=\:\int{vdt} \\ $$$$\:\:\:=\:\int\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}\:+\:\mathrm{97}\right){dt} \\ $$$$\:\:\:=\:{t}^{\mathrm{3}} −\mathrm{10}{t}^{\mathrm{2}} \:+\:\mathrm{97}{t}\:+\:{c} \\ $$$${s}\:=\:\mathrm{0}\:{at}\:{t}=\:\mathrm{3} \\ $$$${solve}\:{and}\:{substitude}\:{for}\:{t}\:=\:\mathrm{11}{s} \\ $$
Commented by Learner-123 last updated on 28/Sep/19
But then you will get displacement not distance.
$${But}\:{then}\:{you}\:{will}\:{get}\:{displacement} \\ $$$${not}\:{distance}. \\ $$
Commented by Rio Michael last updated on 29/Sep/19
yeah your right
$${yeah}\:{your}\:{right} \\ $$
Answered by mr W last updated on 29/Sep/19
a=(dv/dt)=6t−20 ∫_2 ^v dv=∫_5 ^t (6t−20)dt v−2=3t^2 −3×5^2 −20(t−5) ⇒v=(ds/dt)=3t^2 −20t+27 ∫_0 ^s ds=∫_3 ^t (3t^2 −20t+27)dt s=t^3 −3^3 −10(t^2 −3^2 )+27(t−3) ⇒s=t^3 −10t^2 +27t−18 at t=0: s=−18 at t=11: s=11^3 −10×11^2 +27×11−18=400 Δs=400−(−18)=418 m distance travelled=418 m
$${a}=\frac{{dv}}{{dt}}=\mathrm{6}{t}−\mathrm{20} \\ $$$$\int_{\mathrm{2}} ^{{v}} {dv}=\int_{\mathrm{5}} ^{{t}} \left(\mathrm{6}{t}−\mathrm{20}\right){dt} \\ $$$${v}−\mathrm{2}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{3}×\mathrm{5}^{\mathrm{2}} −\mathrm{20}\left({t}−\mathrm{5}\right) \\ $$$$\Rightarrow{v}=\frac{{ds}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}+\mathrm{27} \\ $$$$\int_{\mathrm{0}} ^{{s}} {ds}=\int_{\mathrm{3}} ^{{t}} \left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}+\mathrm{27}\right){dt} \\ $$$${s}={t}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} −\mathrm{10}\left({t}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)+\mathrm{27}\left({t}−\mathrm{3}\right) \\ $$$$\Rightarrow{s}={t}^{\mathrm{3}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{27}{t}−\mathrm{18} \\ $$$${at}\:{t}=\mathrm{0}:\:{s}=−\mathrm{18} \\ $$$${at}\:{t}=\mathrm{11}:\:{s}=\mathrm{11}^{\mathrm{3}} −\mathrm{10}×\mathrm{11}^{\mathrm{2}} +\mathrm{27}×\mathrm{11}−\mathrm{18}=\mathrm{400} \\ $$$$\Delta{s}=\mathrm{400}−\left(−\mathrm{18}\right)=\mathrm{418}\:{m} \\ $$$${distance}\:{travelled}=\mathrm{418}\:{m} \\ $$
Commented by Learner-123 last updated on 29/Sep/19
thanks sir.
$${thanks}\:{sir}. \\ $$

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