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 ofher wheels is 35 cm and the bike moves at 70p cm/sec in the direction indicated. As the bike moves from left to right in the picture below, the chalk mark will traceout a path in the imposed coordinate system:


The parametric equations for the path of the chalk mark are
x(t) = 70pt + 35cos(–2pt – p/2)
y(t) = 35 + 35sin(–2pt – p/2)
A portion of the path (on the time interval [0,2]) is shown in this figure:
Compute x'(t) at these times:
t = 0sec =?
t = 0.25sec =?
t = 0.5sec = ?
t = 0.75sec = ?
t = 1 sec =?
Compute y'(t) at these times:
t = 0.25sec
t = 0.5sec
t = 0.75sec
t = 1 sec

Answer 1

(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?))/?