A-particle-moves-along-a-straight-line-such-that-its-velocity-v-m-s-1-is-given-by-v-t-3-4t-2-3t-where-t-is-time-in-seconds-after-passing-through-fixed-point-O-Find-the-total-distance-in

Question Number 114145 by ZiYangLee last updated on 17/Sep/20

A particle moves along a straight line

such that its velocity, v m s^{- 1}, is given

by v = t^{3} - 4 t^{2} + 3 t, where t is time, in

seconds, after passing through fixed

point O .

Find the total distance, in m, travelled

by the particle until the particle returned

to the fixed point O for the second time .

Answered by mr W last updated on 17/Sep/20

s (t) = \int_{0}^{t} v d t = \frac{t^{4}}{4} - \frac{4 t^{3}}{3} + \frac{3 t^{2}}{2}

s (t) = \frac{t^{4}}{4} - \frac{4 t^{3}}{3} + \frac{3 t^{2}}{2}

s_{m a x} i s w h e n v = 0,

t^{3} - 4 t^{2} + 3 t = 0

(t^{2} - 4 t + 3) t = 0

(t - 1) (t - 3) t = 0

t = 0

t = 1

t = 3

a t t = 1 : s_{m a x} = \frac{1^{4}}{4} - \frac{4 \times 1^{3}}{3} + \frac{3 \times 1^{2}}{2} = \frac{5}{12}

t o t a l d i s t a n c e t r a v e l l e d :

s = 2 \times s_{m a x} = \frac{5}{6} m

Commented by ZiYangLee last updated on 17/Sep/20

oops sorry the answer is 5 . 335 m instead

Commented by mr W last updated on 17/Sep/20

{i t}^{'} s u p o n w h a t i s m e a n t w i t h “ r e t u r n

t o p o i n t O f o r t h e s e c o n d {t i m e}^{″} .

w h a t i m e a n t i s t h a t t h e o b j e c t l e a v e s

p o i n t O a n d c o m e s b a c k . b u t i f i t

m e a n s t h a t t h e o b j e c t l e a v e s p o i n t O

a n d c o m e s b a c k a n d l e a v e s a g a i n a n d

c o m e s b a c k a g a i n, t h e n t h e r e s u l t i s :

a t t = 3 : s_{m i n} = \frac{3^{4}}{4} - \frac{4 \times 3^{3}}{3} + \frac{3 \times 3^{2}}{2} = - \frac{9}{4}

\Rightarrow t o t a l d i s t a n c e m o v e d :

2 \times \frac{5}{12} + 2 \times \frac{9}{4} = \frac{16}{3} = 5.333 m

Commented by mr W last updated on 17/Sep/20

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