Math, asked by tarushiagarwal2123, 9 months ago

- A pedestrian left point A for a walk, going with
the speed of 5 km/h. When the pedestrian was at
a distance of 6 km from A, a cyclist followed him,
starting from A and cycling at a speed 9 km/h higher
than that of the pedestrian. When the cyclist overtook
the pedestrian, they turned back and returned to A
together, at the speed of 4 km/h. At what y will the
time spent by the pedestrian on his total journey from
A to A be the least?
(a) 5 km/h
(b) 6 km/h
(c) 6.1 kmph
(d) 5.5 km/h​

Answers

Answered by RvChaudharY50
1

Given :- A pedestrian left point A for a walk . When the pedestrian was at a distance of 6 km from A, a cyclist followed him, starting from A and cycling at a speed 9 km/h higher than that of the pedestrian. When the cyclist overtook the pedestrian, they turned back and returned to A together, at the speed of 4 km/h. At what speed will the time spent by the pedestrian on his total journey from A to A be the least ?

(a) 5 km/h

(b) 6 km/h

(c) 6.1 kmph

(d) 5.5 km/h

Answer :-

Let us assume that, the speed of pedestrian is x km/h .

so,

→ speed of cyclist = (x + 9) km/h .

given that, the cyclist starts from A after pedestrian covers 6 km. Let after y hours he cross the pedestrian .

so,

→ Distance travelled by pedestrian = (6 + xy) km..

→ Distance travelled by cyclist = (x + 9)y = (xy + 9y) km .

then,

→ 6 + xy = xy + 9y

→ 9y = 6

→ y = (2/3) hours.

now,

→ Total distance travelled by pedestrian from A = 6 + (2x/3) km.

then,

→ Time taken by pedestrian to cover this distance = D/S = [6 + (2x/3)]/x = {(6/x) + (2/3)} hours. -------- Eqn.(1)

given that, while returning there speed were 4km/h .

so,

→ Time taken by pedestrian to reach back to A = D/S = [6 + (2x/3)]/4 = {(6/4) + (2x/12)} = {(3/2) + (x/6)} hours -------- Eqn.(2)

we have to find value of x for which total time of pedestrian from A to A would be minimum .

so, using Eqn.(1) and Eqn.(2)

→ Total time taken by pedestrian from A to A = {(6/x) + (2/3)} + {(3/2) + (x/6)} = (6/x) + (x/6) + (2/3) + (3/2) = (6/x) + (x/6) + (4+9)/6 = {(6/x) + (x/6) + (13/6)} hours.

checking options for minimum value now, we get,

(a) 5km/h

→ (6/5) + (5/6) + (13/6)

→ (6/5) + (18/6)

→ 1(1/5) + 3

→ 4(1/5) hours = 4 hours + 12 minutes .

(b) 6 km/h

→ (6/6) + (6/6) + (13/6)

→ 1 + 1 + 2(1/6)

→ 4(1/6) hours = 4 hours + 10 minutes .

(c) 6.1 km/h

→ (6/6.1) + (6.1/6) + (13/6)

→ (6/6.1) + (19.1/6)

→ (60/61) + (191/60)

→ (3600 + 11651)/3660

→ (15251/3660)

→ 4(611/3660) ≈ 4 hours + greater than 10 minutes .

(d) 5.5 km/h

→ (6/5.5) + (5.5/6) + (13/6)

→ (60/55) + (18.5/6)

→ (12/11) + (37/12)

→ (144 + 407)/132

→ (551/132)

→ 4(23/132) = 4 hours + greater than 10 minutes .

therefore, we can conclude that, at 6km/h (B) the time spent by pedestrian will be minimum .

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Answered by amitnrw
2

Given :  A pedestrian left point A for a walk, going with the speed of y km/h.  

When the pedestrian was at a distance of 6 km from A, a cyclist followed him, starting from A and cycling at a speed 9 km/h higher than that of the pedestrian. When the cyclist overtook the pedestrian, they turned back and returned to A together, at the speed of 4 km/h.

To Find : At what y will the time spent by the pedestrian on his total journey from A to A be the least?

(a) 5 km/h

(b) 6 km/h

(c) 6.1 kmph

(d) 5.5 km/h​

Solution:

pedestrian speed = y km/hr

cyclist speed =  y + 9 km/hr

Cyclist relative speed = y + 9 - y =  9 km/hr

cyclist relative distance  =  6 km/hr

Time taken  to overtake =  6/9  = 2/3  hrs

Travel time of pedestrian from A to return point  = time to cross 6 km + 2/3 hrs   =  6/y   + 2/3  hrs

Distance covered  =  6  + (2/3)y   km

Return speed = 4 km/hr

Return time  =  ( 6 + (2/3)y)/4  = 3/2  +  y/6  hrs

Total travel time T of Pedestrian from A to A  = 6/y  + 2/3  +  3/2  +  y/6

T = 6/y  + y/6  + 13/6

dT/dy  = -6/y²  + 1/6

dT/dy  =   0

=> -6/y²  + 1/6

 -6/y²  + 1/6 = 0

=> y² = 36

=> y = 6  

d²T/dy²  = 12/y³  > 0   Hence Time is minimum  at y = 6

At 6 km/hr  the the time spent by the pedestrian on his total journey from

A to A be the least  

Hence  6 km/hr  is correct option.

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