7. When standing 2.0 m in front of an amusement park mirror, you notice that your image is three times taller. What is the radius of curvature of the mirror?

nswer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}

s

1

+

s

′

1

=

R

2

\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}

2.0m

1

+

s

′

1

=

1.50m

2

Solve s'

\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}

s

′

1

=

1.50m

2

−

2.0m

1

\frac{1}{s'} = 1.833 m

s

′

1

= 1.833m

s' = - 0.545 ms

′

=−0.545m

To determine the speed of the trick to the highway

\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}

\frac{ds}{dt} = 26.88 m/s

dt

ds

=26.88m/s

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

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