A bicyclist is interested in the motion of a point on the tread of her tire, so

Question

A bicyclist is interested in the motion of a point on the tread of her tire, so she marks the tread at one point with chalk. She then rides the bike along a flat, level road. She imposes a coordinate system as shown in the figure below such that the chalk mark is touching the ground at the point (0,0) at time t=0. The radius of her wheels is 35 cm and the bike moves at 70 cm/sec in the direction indicated. As the bike moves from left to right in the picture below, the chalk mark will trace out a path in the imposed coordinate system:

The parametric equations for the path of the chalk mark are
x(t) = 70t + 35cos(–2t – /2)
y(t) = 35 + 35sin(–2t – /2)
A portion of the path (on the time interval [0,2]) is shown in this figure:

Give EXACT ANSWERS to all questions below; use pi to denote the number .

(a) The horizontal velocity is  .
(b) The vertical velocity is  .
(c) Compute x'(t) at these times:
t = 0sec  
t = 0.25sec  
t = 0.5sec  
t = 0.75sec  
t = 1 sec  

(d) Compute y'(t) at these times:
t = 0sec  
t = 0.25sec  
t = 0.5sec  
t = 0.75sec  
t = 1 sec  

(e) When is the first time the horizontal velocity will be a maximum?  
(f) When is the first time the horizontal velocity will be a minimum?  
(g) What is the maximum horizontal velocity?  
(h) What is the minimum horizontal velocity?  
(i) When is the first time the vertical velocity will be a maximum?  
(j) When is the first time the vertical velocity be a minimum?  
(k) What is the maximum vertical velocity?  
(l) What is the minimum vertical velocity?  
(m) The speed of the moving chalk mark at time t is defined by the equation
s(t) = {(x'(t))2 + (y'(t))2}.
Compute s(t).  
(n) During the first 2 seconds, how many times will the speed be equal to 50 cm/sec?  

Explanation / Answer

(a)Given x(t) = 70?t + 35cos(–2?t – ?/2)

Horizontal velocity=v(t)=d(x(t))/dt=70?+70?sin(–2?t – ?/2)

(b) Given y(t) = 35 + 35sin(–2?t – ?/2)

Vetical velocity v(t)=d{y(t)}/dt=-70?cos(-2?t-?/2)

(c) At t=0 x'(0)=70?+70?(-1)=0

AT t=0.25, x'(0.25)=70?+70?sin(-?)=70?

at t=0.5, x'(0.5)=70?+70?sin(-3?/2)=70?+70?=140?

at t=0.75, x'(0.75)=70?+70?sin(-2?)=70?

at t=1, x'(1)=70?+70?sin(-5?/2)=0

(d)y'(0)=70?cos(-?/2)=0

y'(0.25)=70?cos(-?)=-70?

v(0.5)=70?cos(-3?/2)=0

y'(0.75)=70?cos(-2?)=70?

y'(1)=70?cos(-5?/2)=0

(e)The horizontal velocity maximum At t=0.5

(f)The horizontal velocity mininum at t=0 and t=1

(g)Maximum horizontal velocity=140?

(h) The minimum horizontal velocity=0

(i) the maximum verical velocity at t=0.75

(j) the minimum vertical velocity at t=0.25

(k) The maximimum vertical velocity v(0.75)=70?

(l)the minimum vertical velocity v(0.25)=-70?

(m)s(t) = ?{(x'(t))2 + (y'(t))2}

s(t)=?{{70?+70?sin(-2?t-?/2)}^2+{70?cos(-2?t-?/2)}^2}=140?|in(?t)|

(n)when 140?sin(?t)=50 cm/sec

or, sin(?t)=50/140?=5/(14?)

or, t=(arcsin(5/14?))/?

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