PHYSICS HOMEWORK HELP PLEASE?

Decide what units to use. I'll stick with miles and hours.

12s = 12/3600 h = 1/300 h

Original speed is v.

Time to travel 1 mile is t = distance/speed = 1/v

Increased speed is v' = v + 7.3

Time to travel 1 mile is t' = distance/speed = 1/v' = 1/(v + 7.3)

The time to go 1 mile has decreased by 1/300 hours:

t - t' = 1/300

1/v - 1/(v + 7.3) = 1/300

Multiply through by 300v(v + 7.3)

300(v + 7.3) - 300v = v(v + 7.3)

300v + 2190 - 300v = v² + 7.3v

v² + 7.3v - 2190 = 0

Solve the quadratic equation to give v = 43.3 or v = -50.6

A negative speed is not meaningful, so the original speed was 43.3mi/h.

let the original speed be 'v' mph

distance = speed x time

let 't' be in hours (therefore 12 s = 12/3600 hrs)

=> 1 mile = vt ---------------------------------> (i)

also on speed increase:

1 mile = (v + 7.3)[t - (12/3600] ---------------------------------> (ii)

or 1 = vt - (12/3600)v + 7.3t - (87.6/3600)

from equation (i) => 1 = 1 - 0.0033v + 7.3t - 0.02433

=> 0.0033v = 7.3t - 0.02433

& 0.0033vt = 0.0033 = (7.3t - 0.02433)t

=> 7.3t² - 0.02433t - 0.0033 = 0

solving => t = 0.02299 hrs

therefore v = 1/t = 43.497 mph

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or

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if we use miles per second as the speed unit: 7.3 mph = 0.0020277 mps

=> 1 mile = vt ---------------------------------> (i)

also on speed increase:

1 mile = (v + 0.0020277)[t - 12] ---------------------------------> (ii)

or 1 = vt - (12)v + 0.0020277t - (0.02433)

from equation (i) => 1 = 1 - 12v + 0.0020277t - 0.02433

=> 12v = 0.0020277t - 0.02433

& 12vt = 12 = (0.0020277t - 0.02433

)t

=> 0.0020277t² - 0.02433t - 12 = 0

solving => t = 83.11097 s

therefore v = 1/t = 0.01203210 mps ~= 0.0120321 x 3600 = 43.316 mph

hope this helps

(original speed) * t = 1 mile

(original speed + 7.3 mph) (t - 12s) = 1 mile

...

algebra

....

original speed = sqrt[(7.3/12) mile^2/hour/sec - (3.65 mph)^2] - 3.65 mph

I usually just throw those sorts of expressions into google, but for school you should probably do the seconds-to-hours conversion yourself.