Menu Close

A-particle-in-a-straight-line-with-uniform-deceleration-has-a-velocity-of-40ms-1-at-point-P-20ms-1-at-point-Q-and-comes-to-rest-at-point-R-where-QR-50m-calculate-the-i-Distance-PQ-ii-T

Tinku Tara 0 Comments
Pin




Question Number 132683 by otchereabdullai@gmail.com last updated on 15/Feb/21
A particle in a straight line with uniform deceleration has a velocity of 40ms^(−1) at point P , 20ms^(−1) at point Q and comes to rest at point R where QR= 50m. calculate the (i) Distance PQ (ii)Time taken to cover PQ (iii)Time taking to cover PR
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with} \\ $$$$\mathrm{uniform}\:\mathrm{deceleration}\:\mathrm{has}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of} \\ $$$$\mathrm{40ms}^{−\mathrm{1}} \:\mathrm{at}\:\mathrm{point}\:\mathrm{P}\:,\:\mathrm{20ms}^{−\mathrm{1}} \:\mathrm{at}\:\mathrm{point}\:\mathrm{Q} \\ $$$$\mathrm{and}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{point}\:\mathrm{R}\:\mathrm{where}\: \\ $$$$\mathrm{QR}=\:\mathrm{50m}.\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Distance}\:\mathrm{PQ} \\ $$$$\left(\mathrm{ii}\right)\mathrm{Time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{PQ} \\ $$$$\left(\mathrm{iii}\right)\mathrm{Time}\:\mathrm{taking}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{PR} \\ $$$$ \\ $$
Answered by mr W last updated on 16/Feb/21
Method I say uniform deceleration = a=constant a=(dv/dt)=(dv/ds)×(ds/dt)=v(dv/ds) vdv=ads ∫vdv=a∫ds ⇒(v^2 /2)+C=as at P: s=0, v=40 ⇒((40^2 )/2)+C=0 ...(i) at Q: s=x, v=20 ⇒((20^2 )/2)+C=ax ...(ii) at R: s=x+50, v=0 ⇒C=a(x+50) ...(iii) ⇒C=−800 ⇒ax=200−800=−600 ⇒a(x+50)=−800 ((x+50)/x)=((800)/(600)) ((x+50)/x)=(4/3) ⇒x=150 ⇒a=−((600)/(150))=−4 m/s^2 time for PQ=t_1 : v_Q =v_P +at_1 t_1 =((20−40)/(−4))=5 s time for QR=t_2 : t_2 =((0−20)/(−4))=5 s
$${Method}\:{I} \\ $$$${say}\:{uniform}\:{deceleration}\:=\:{a}={constant} \\ $$$${a}=\frac{{dv}}{{dt}}=\frac{{dv}}{{ds}}×\frac{{ds}}{{dt}}={v}\frac{{dv}}{{ds}} \\ $$$${vdv}={ads} \\ $$$$\int{vdv}={a}\int{ds} \\ $$$$\Rightarrow\frac{{v}^{\mathrm{2}} }{\mathrm{2}}+{C}={as} \\ $$$${at}\:{P}:\:{s}=\mathrm{0},\:{v}=\mathrm{40}\:\Rightarrow\frac{\mathrm{40}^{\mathrm{2}} }{\mathrm{2}}+{C}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$${at}\:{Q}:\:{s}={x},\:{v}=\mathrm{20}\:\Rightarrow\frac{\mathrm{20}^{\mathrm{2}} }{\mathrm{2}}+{C}={ax}\:\:\:…\left({ii}\right) \\ $$$${at}\:{R}:\:{s}={x}+\mathrm{50},\:{v}=\mathrm{0}\:\Rightarrow{C}={a}\left({x}+\mathrm{50}\right)\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow{C}=−\mathrm{800} \\ $$$$\Rightarrow{ax}=\mathrm{200}−\mathrm{800}=−\mathrm{600} \\ $$$$\Rightarrow{a}\left({x}+\mathrm{50}\right)=−\mathrm{800} \\ $$$$\frac{{x}+\mathrm{50}}{{x}}=\frac{\mathrm{800}}{\mathrm{600}} \\ $$$$\frac{{x}+\mathrm{50}}{{x}}=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow{x}=\mathrm{150} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{600}}{\mathrm{150}}=−\mathrm{4}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$${time}\:{for}\:{PQ}={t}_{\mathrm{1}} : \\ $$$${v}_{{Q}} ={v}_{{P}} +{at}_{\mathrm{1}} \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{20}−\mathrm{40}}{−\mathrm{4}}=\mathrm{5}\:{s} \\ $$$${time}\:{for}\:{QR}={t}_{\mathrm{2}} : \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{0}−\mathrm{20}}{−\mathrm{4}}=\mathrm{5}\:{s} \\ $$
Commented by mr W last updated on 15/Feb/21
Commented by otchereabdullai@gmail.com last updated on 15/Feb/21
Prof W thank you soooooo much!
$$\mathrm{Prof}\:\mathrm{W}\:\:\mathrm{thank}\:\mathrm{you}\:\mathrm{soooooo}\:\mathrm{much}! \\ $$
Answered by mr W last updated on 16/Feb/21
Method II let the particle move backwards from R to Q to P. it moves then with an uniform acceleration a. s_1 =RQ=50 v_Q ^2 −v_R ^2 =2as_1 ⇒a=((20^2 −0)/(2×50))=4 m/s^2 time from R to Q=t_1 v_Q =v_R +at_1 ⇒t_1 =((20−0)/4)=5 s s_2 =QP=x v_P ^2 −v_Q ^2 =2ax ⇒x=((40^2 −20^2 )/(2×4))=150 m time from Q to P=t_2 ⇒t_2 =((40−20)/4)=5 s
$${Method}\:{II} \\ $$$$ \\ $$$${let}\:{the}\:{particle}\:{move}\:{backwards}\:{from} \\ $$$${R}\:{to}\:{Q}\:{to}\:{P}.\:{it}\:{moves}\:{then}\:{with}\:{an} \\ $$$${uniform}\:{acceleration}\:{a}. \\ $$$${s}_{\mathrm{1}} ={RQ}=\mathrm{50} \\ $$$${v}_{{Q}} ^{\mathrm{2}} −{v}_{{R}} ^{\mathrm{2}} =\mathrm{2}{as}_{\mathrm{1}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{20}^{\mathrm{2}} −\mathrm{0}}{\mathrm{2}×\mathrm{50}}=\mathrm{4}\:{m}/{s}^{\mathrm{2}} \\ $$$${time}\:{from}\:{R}\:{to}\:{Q}={t}_{\mathrm{1}} \\ $$$${v}_{{Q}} ={v}_{{R}} +{at}_{\mathrm{1}} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{20}−\mathrm{0}}{\mathrm{4}}=\mathrm{5}\:{s} \\ $$$${s}_{\mathrm{2}} ={QP}={x} \\ $$$${v}_{{P}} ^{\mathrm{2}} −{v}_{{Q}} ^{\mathrm{2}} =\mathrm{2}{ax} \\ $$$$\Rightarrow{x}=\frac{\mathrm{40}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}}=\mathrm{150}\:{m} \\ $$$${time}\:{from}\:{Q}\:{to}\:{P}={t}_{\mathrm{2}} \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\frac{\mathrm{40}−\mathrm{20}}{\mathrm{4}}=\mathrm{5}\:{s} \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Feb/21
God bless you prof W
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{W} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *