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Question Number 57735 by Tawa1 last updated on 10/Apr/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Apr/19
	$s=2t^3 −13t^2 +20t s=t(2t^2 −13t+20) s=t(2t^2 −8t−5t+20) =t{2t(t−4)−5(t−4)} =t(t−4)(2t−5) s=0 when t=0,t=4 and t=2.5 when t=4 s=0 that means in 4 seconds it cover same distance twice but in opposite direction .hence displace ment is zero. S_(t=2) =2(2−4)(2×2−5)=2×−2×−1=4 so total distance covered in 4 seconds=2×4=8 meter S=2t^3 −13t^2 +20t (dS/dt)=6t^2 −26t+20 (d^2 s/dt^2 )=12t−26 given accelaration −ve (d^2 s/dt^2 )<0 12t−26<0 →t<((26)/(12)) so in time duration ((13)/6)> t>0 acc is negetive pls check is it correect answer.. then maximum speed to[be calculated (ds/dt)=6t^2 −26t+20 when t=0 (ds/dt)=20 that means initial velocity=20. now (d^2 s/dt^2 )=12t−26 now in time interval ((26)/(12))>t>0 accelaration is −ve so in time interval [0,((26)/(12))] vel_(max) =speed_(max) =20 in time interval [((26)/(12)),4] acc is +ve (ds/dt)=6t^2 −26t+20 6t^2 −26t+20=0 3t^2 −13t+10=0 3t^2 −3t−10t+10=0 3t(t−1)−10(t−1)=0 (t−1)((3t−10)=0 when t=1 velocity=0 when t=((10)/3) velocity=0 so in time interval 4>t>0 max speed is=20 pls check$
	$s = 2 t^{3} - 13 t^{2} + 20 t$ $s = t (2 t^{2} - 13 t + 20)$ $s = t (2 t^{2} - 8 t - 5 t + 20)$ $= t {2 t (t - 4) - 5 (t - 4)}$ $= t (t - 4) (2 t - 5)$ $s = 0 w h e n t = 0, t = 4 a n d t = 2.5$ $w h e n t = 4 s = 0$ $t h a t m e a n s i n 4 s e c o n d s i t c o v e r s a m e d i s t a n c e$ $t w i c e b u t i n o p p o s i t e d i r e c t i o n . h e n c e d i s p l a c e$ $m e n t i s z e r o .$ $S_{t = 2} = 2 (2 - 4) (2 \times 2 - 5) = 2 \times - 2 \times - 1 = 4$ $s o t o t a l d i s t a n c e c o v e r e d i n 4 s e c o n d s = 2 \times 4 = 8 m e t e r$ $S = 2 t^{3} - 13 t^{2} + 20 t$ $\frac{d S}{d t} = 6 t^{2} - 26 t + 20$ $\frac{d^{2} s}{{d t}^{2}} = 12 t - 26$ $g i v e n a c c e l a r a t i o n - v e$ $\frac{d^{2} s}{{d t}^{2}} < 0$ $12 t - 26 < 0 \to t < \frac{26}{12}$ $s o i n t i m e d u r a t i o n \frac{13}{6} > t > 0 a c c i s n e g e t i v e$ $p l s c h e c k i s i t c o r r e e c t a n s w e r . .$ $t h e n m a x i m u m s p e e d t o [b e c a l c u l a t e d$ $\frac{d s}{d t} = 6 t^{2} - 26 t + 20 w h e n t = 0 \frac{d s}{d t} = 20 t h a t m e a n s i n i t i a l$ $v e l o c i t y = 20 .$ $n o w \frac{d^{2} s}{{d t}^{2}} = 12 t - 26 n o w i n t i m e i n t e r v a l$ $\frac{26}{12} > t > 0 a c c e l a r a t i o n i s - v e$ $s o i n t i m e i n t e r v a l [0, \frac{26}{12}] {v e l}_{m a x} = {s p e e d}_{m a x} = 20$ $i n t i m e i n t e r v a l [\frac{26}{12}, 4] a c c i s + v e$ $\frac{d s}{d t} = 6 t^{2} - 26 t + 20$ $6 t^{2} - 26 t + 20 = 0$ $3 t^{2} - 13 t + 10 = 0$ $3 t^{2} - 3 t - 10 t + 10 = 0$ $3 t (t - 1) - 10 (t - 1) = 0$ $(t - 1) ((3 t - 10) = 0$ $w h e n t = 1 v e l o c i t y = 0$ $w h e n t = \frac{10}{3} v e l o c i t y = 0$ $s o i n t i m e i n t e r v a l 4 > t > 0$ $m a x s p e e d i s = 20$ $p l s c h e c k$
	Commented by Tawa1 last updated on 11/Apr/19

	$God bless you sir . It should be right sir . I will study your solution .$
	Commented by Tawa1 last updated on 11/Apr/19

	$God bless you sir . I appreciate your time .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Apr/19
	$3)a) when they meet x_p =x_Q 5t^2 (t+7)=t^3 (t+3) t^2 (5t+35)−t^3 (t+3)=0 t^2 (5t+35−t^2 −3t)=0 when t=0 they meet at gas station −t^2 +2t+35=0 t^2 −2t−35=0 t^2 −7t+5t−35=0 t(t−7)+5(t−7)=0 (t−7)(t+5)=0 t≠−5 so after t=7 they meet. b)v_p =(dx_p /dt)=((d(5t^3 +35t^2 ))/dt)=15t^2 +70t v_Q =((d(t^4 +3t^3 ))/dt) =4t^3 +9t^2 so velocity of p w.r.t Q=v_p −v_Q =(15t^2 +70t)−(4t^3 +9t^2 ) =6t^2 +70t−4t^3 c)v_p >v_Q v_p −v_Q >0 (15t^2 +70t)−(4t^3 +9t^2 )>0 6t^2 +70t−4t^3 >0 t(4t^2 −6t−70)<0 2t(2t^2 −3t−35)<0 2t(2t^2 −10t+7t−35)<0 2t{2t(t−5)+7(t−5)}<0 2t(t−5)(2t+7)<0 t≠negetive so[critical value of t =0,5 f(t)=2t(t−5)(2t+7)<0 when t>5 f(t)>0 so when t>0 but t<5 then f(t)<0 so in time interval 5>t>0 v_p >v_Q$
	$3) a) w h e n t h e y m e e t x_{p} = x_{Q}$ $5 t^{2} (t + 7) = t^{3} (t + 3)$ $t^{2} (5 t + 35) - t^{3} (t + 3) = 0$ $t^{2} (5 t + 35 - t^{2} - 3 t) = 0$ $w h e n t = 0 t h e y m e e t a t g a s s t a t i o n$ $- t^{2} + 2 t + 35 = 0$ $t^{2} - 2 t - 35 = 0$ $t^{2} - 7 t + 5 t - 35 = 0$ $t (t - 7) + 5 (t - 7) = 0$ $(t - 7) (t + 5) = 0$ $t \neq - 5 s o a f t e r t = 7 t h e y m e e t .$ $b) v_{p} = \frac{{d x}_{p}}{d t} = \frac{d (5 t^{3} + 35 t^{2})}{d t} = 15 t^{2} + 70 t$ $v_{Q} = \frac{d (t^{4} + 3 t^{3})}{d t} = 4 t^{3} + 9 t^{2}$ $s o v e l o c i t y o f p w . r . t Q = v_{p} - v_{Q}$ $= (15 t^{2} + 70 t) - (4 t^{3} + 9 t^{2})$ $= 6 t^{2} + 70 t - 4 t^{3}$ $c) v_{p} > v_{Q} v_{p} - v_{Q} > 0$ $(15 t^{2} + 70 t) - (4 t^{3} + 9 t^{2}) > 0$ $6 t^{2} + 70 t - 4 t^{3} > 0$ $t (4 t^{2} - 6 t - 70) < 0$ $2 t (2 t^{2} - 3 t - 35) < 0$ $2 t (2 t^{2} - 10 t + 7 t - 35) < 0$ $2 t {2 t (t - 5) + 7 (t - 5)} < 0$ $2 t (t - 5) (2 t + 7) < 0$ $t \neq n e g e t i v e s o [c r i t i c a l v a l u e o f t = 0, 5$ $f (t) = 2 t (t - 5) (2 t + 7) < 0$ $w h e n t > 5 f (t) > 0$ $s o w h e n t > 0 b u t t < 5 t h e n f (t) < 0$ $s o i n t i m e i n t e r v a l 5 > t > 0$ $v_{p} > v_{Q}$
	Commented by Tawa1 last updated on 11/Apr/19

	$Wow, God bless you sir . Thanks for your time sir .$
	Commented by peter frank last updated on 13/Apr/19

	$t h a n k y o u$
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