Question
Asked Oct 2, 2019
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Jane and Bob see each other when 100 m apart. They are moving toward each other, Jane at speed 2.8m/s and Bob at speed 3.4m/s with respect to the ground.

 

Find the time elapsed for Bob and Jane to meet up.

Find the distance Jane moves before meeting up with Bob.

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Expert Answer

Step 1

Assume that the Jane and Bob are moving along the x-axis

Assume that Jane's starting point is at xoj = 0 m and Bob's starting point is at xB = 100 m. Also
assume Jane and Bob meets at distance x from the origin
The expression for the final position of Jane is given by,
Here, is the point where Jane and Bob meets, v is the velocity of Jane and t is the time elapsed
before they meet
The expression for the final position of Bob is given by
х %3D Хов + 1в
Here, s the velocity of Bob
(2)
Equate equations (1) and (2)
OB
t(v-VB)x0B -xo
=
Хов — Хо
OJ
t =
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Assume that Jane's starting point is at xoj = 0 m and Bob's starting point is at xB = 100 m. Also assume Jane and Bob meets at distance x from the origin The expression for the final position of Jane is given by, Here, is the point where Jane and Bob meets, v is the velocity of Jane and t is the time elapsed before they meet The expression for the final position of Bob is given by х %3D Хов + 1в Here, s the velocity of Bob (2) Equate equations (1) and (2) OB t(v-VB)x0B -xo = Хов — Хо OJ t =

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Step 2

Since Bob moves towards opposite to the positive x-direction, his velocity wi...

100 m-0 m
t =
2.8 m/s-(3.4 m/s)
100 m
6.2 m/s
=16.13 s
To find the distance moved by Jane before meeting up with Bob, substitute the obtained value oft in
equation (1)
x 0 m(2.8 m/s) (16.13 s)
= 45.16 m
help_outline

Image Transcriptionclose

100 m-0 m t = 2.8 m/s-(3.4 m/s) 100 m 6.2 m/s =16.13 s To find the distance moved by Jane before meeting up with Bob, substitute the obtained value oft in equation (1) x 0 m(2.8 m/s) (16.13 s) = 45.16 m

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