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Question

The position vector r=5.00t^i+(et+ft2)^j locates a particle as a function of time t. Vector r is in meters, t is in seconds, and factors e and f are constants. Below Diagram gives the angle θ of the particle's direction of travel as a function of t(θ) is measured from the positive x direction). What are 'e' and 'f', including units?


A

e=3563 m/s, f=5243 m/s2

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B

e=3563 m/s, f=5243 m/s

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C

e=3563 m/s, f=5243 m/s2

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D

e=3563 m/s, f=5243 m/s2

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Solution

The correct option is A

e=3563 m/s, f=5243 m/s2


Given the graph between θ and t you can write the equation of line using straight line equation,
y=mx+c
θ=52t+35
Given that this θ is the particles direction of travel.
We know direction of travel is the direction of velocity vector and not position vector.
So let's find velocity by differentiating
r=5t^i+(et+ft2)^j
v=5^i+(e+2ft)^j
vx=5 m/svy=(e+2ft)m/s.

Now tan θ will always be vyvx

From the graph, we can see that
At t=14, θ=0,tan(θ)=0

att=14s, vyvx=0 vy=0
vy=e+2ft
e+2f(14)=0
e+28f=0 ......(1)
at t= 2 s, θ=30
24f=53
f=5243
e=28f=28×(5243)
=76×53=3563

To find units of e and f, we know
vy=(e+2ft)m/s
e must have a unit of m/s and ft must have a unit of m/s.
f must have a unit of m/s2 as t has a unit of 's'

e=3563 m/s and f=5243 m/s2


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