Need help calculating distance by a car

[Guest]

Cop says client was going 55mph on a road.

Client said she was going 40 mph

Client said she was at a red light and then accelerated so basically 0-40mph. She gave me the coordinates of the road.

So client went 0-40 in 528 feet (when cop saw her)

Cop says she was going 55mph so essentially 0-55mph in 528 feet.

My question is cops version possible, she drives a budget 2015 kia optima which goes 0-60 in 7.4-8.4 seconds per specs...im trying to call the officer a liar by way of saying what he saw was an impossibility or really unlikely.

[Guest]

Good morning.

Accordingly to specs of the car (I didn't check it, ok?), the cop could be right, at all!

Acceleration of the car: 0-60 mph in 7,4-8,4s, right?
Converting to feet per second (fps) = [tex]60 \cdot \dfrac{ 5280\text{feet}}{3600\text{seconds}}=88\; ft/s[/tex]

Calculating two accelerations (7,4 and 8,4 seconds):
[tex]a_1=\dfrac{88}{7,4}=11,89ft/s^2\\a_2=\dfrac{88}{8,4}=10,48ft/s^2[/tex]

These are the specs acceleration.

Now, let's calc the car acceleration during 528 feet.

If the car gets 55mph = [tex]55\cdot\dfrac{5280}{3600}\approx 80,67[/tex]
Using the following equation:
[tex]v^2=v_0^2+2a\Delta s\\
80,67^2=0^2+2a\cdot 528\\
6\,507,6489=1\,056a\\
a=\dfrac{6\,507,6489}{1\,056}\\
a\approx 6,16ft/s^2[/tex]

Totally possible!

In fact, the car could reach:
[tex]v^2=v_0^2+2a\Delta s\\
v^2=0^2+2(10,48)(528)\\
v^2=11\,066,88\\
v\approx 105,20ft/s=72,73mph[/tex]

Hope have helped!

[Baltuilhe]

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