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

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Question Number 132683 by otchereabdullai@gmail.com last updated on 15/Feb/21

$$\mathrm{A}\mathrm{particle}\mathrm{in}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{with}$$$$\mathrm{uniform}\mathrm{deceleration}\mathrm{has}\mathrm{a}\mathrm{velocity}\mathrm{of}$$$${40\mathrm{ms}}^{-1}\mathrm{at}\mathrm{point}\mathrm{P},{20\mathrm{ms}}^{-1}\mathrm{at}\mathrm{point}\mathrm{Q}$$$$\mathrm{and}\mathrm{comes}\mathrm{to}\mathrm{rest}\mathrm{at}\mathrm{point}\mathrm{R}\mathrm{where}$$$$\mathrm{QR}=50\mathrm{m}.\mathrm{calculate}\mathrm{the}$$$$\left(\mathrm{i}\right)\mathrm{Distance}\mathrm{PQ}$$$$\left(\mathrm{ii}\right)\mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{cover}\mathrm{PQ}$$$$\left(\mathrm{iii}\right)\mathrm{Time}\mathrm{taking}\mathrm{to}\mathrm{cover}\mathrm{PR}$$$$$$

Answered by mr W last updated on 16/Feb/21

$$MethodI$$$$sayuniformdeceleration=a=constant$$$$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}$$$$vdv=ads$$$$\int vdv=a\int ds$$$$\Rightarrow \frac{v^2}{2}+C=as$$$$atP:s=0,v=40\Rightarrow \frac{{40}^2}{2}+C=0\dots \left(i\right)$$$$atQ:s=x,v=20\Rightarrow \frac{{20}^2}{2}+C=ax\dots \left(ii\right)$$$$atR:s=x+50,v=0\Rightarrow C=a(x+50)\dots \left(iii\right)$$$$\Rightarrow C=-800$$$$\Rightarrow ax=200-800=-600$$$$\Rightarrow a(x+50)=-800$$$$\frac{x+50}{x}=\frac{800}{600}$$$$\frac{x+50}{x}=\frac{4}{3}\Rightarrow x=150$$$$\Rightarrow a=-\frac{600}{150}=-4m/s^2$$$$$$$$timeforPQ=t_1:$$$$v_Q=v_P+{at}_1$$$$t_1=\frac{20-40}{-4}=5s$$$$timeforQR=t_2:$$$$t_2=\frac{0-20}{-4}=5s$$

Commented by mr W last updated on 15/Feb/21

Commented by otchereabdullai@gmail.com last updated on 15/Feb/21

$$\mathrm{Prof}\mathrm{W}\mathrm{thank}\mathrm{you}\mathrm{soooooo}\mathrm{much}!$$

Answered by mr W last updated on 16/Feb/21

$$MethodII$$$$$$$$lettheparticlemovebackwardsfrom$$$$RtoQtoP.itmovesthenwithan$$$$uniformaccelerationa.$$$$s_1=RQ=50$$$$v_Q^2-v_R^2=2{as}_1$$$$\Rightarrow a=\frac{{20}^2-0}{2\times 50}=4m/s^2$$$$timefromRtoQ=t_1$$$$v_Q=v_R+{at}_1$$$$\Rightarrow t_1=\frac{20-0}{4}=5s$$$$s_2=QP=x$$$$v_P^2-v_Q^2=2ax$$$$\Rightarrow x=\frac{{40}^2-{20}^2}{2\times 4}=150m$$$$timefromQtoP=t_2$$$$\Rightarrow t_2=\frac{40-20}{4}=5s$$

Commented by otchereabdullai@gmail.com last updated on 16/Feb/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{prof}\mathrm{W}$$

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