Question-57735

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

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Apr/19

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) 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 .

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

t h a n k y o u