- #1

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I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation

x = x + v*t + 1/2 * a *t^2

40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2

t=15, but that's wrong....any ideas?

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- Thread starter physicsss
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- #1

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I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation

x = x + v*t + 1/2 * a *t^2

40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2

t=15, but that's wrong....any ideas?

- #2

Tide

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... where a is the acceleration with appropriate units, of course! :-)

- #3

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I still get t=15 with that equation.

- #4

Tide

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- #5

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95km/h * 1h/60min * 1min/60s * 1000m/1km= 26.389 m/s and

140km/h * 1h/60min * 1min/60s * 1000m/1km= 38.889 m/s.

so 26.389*t+1/2*2.0*t^2=38.889(t+1)

and t=15 s.

- #6

Tide

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How do you know your answer is wrong?

- #7

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The hw is online and when I entered 15s and submitted, it said it was incorrect.

- #8

Chronos

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You are getting distance, not acceleration, from that formula. The original formula Tide gave is correct. You need only solve for time. Keep trying.

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- #9

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Chronos, can you clarify what you said?

- #10

Doc Al

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This isn't quite right for several reasons.Tide said:[tex]95t+\frac{1}{2}at^2=140(t+1)[/tex]

Measuring everything from the moment the speeder passes the police car (at t = 0):

Distance police car travels: [itex]x_p = v_p t + \frac{1}{2}a(t-1)^2[/itex], note that acceleration only begins at t = 1 second, [itex]v_p[/itex] is the initial speed of the police car in m/s.

Distance speeder travels: [itex]x_s = v_s t[/itex], [itex]v_s[/itex] is the speed of the speeder in m/s.

So, when the police car overtakes the speeder:

[tex]v_p t + \frac{1}{2}a(t-1)^2 = v_s t[/tex]

All times measured in

- #11

Tide

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Thanks, Al, I stand corrected!

- #12

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I got 14.4 seconds, is that correct?

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