# Another Motion Quesiton

• May 19th 2009, 07:31 PM
Dickson
Another Motion Quesiton
Margaret has found that by a proper choice of gears she can steadily increase her speed on her bike. One day she sets out and, ten minutes later, she achieves her cruising speed of 30 km/h by increasing steadily. How far did she travel in this time?

No idea where to begin
• May 19th 2009, 07:47 PM
justchillin
I suggest you use the equation of motion.

$V = Vi + at$
$30km/h = 8.333m/s$
$8.333m/s = 0 + a(10*60)$
$a = 0.014m/s^{2}$

Use the other equation of motion

$d =\frac{1}{2}at^{2}$
$d = \frac{1}{2}(0.014)(600)^{2}= 2500meters = 2.5 km$
• May 19th 2009, 07:49 PM
derfleurer
If your acceleration is constant, just use rise/run from (0,0) to (10,30) to calculate it. Then you can just use a distance equation for your first integral of 3/2x^2. Not really a calc question.
• May 19th 2009, 07:53 PM
Dickson
Yeah we're not using distance equations like that though
• May 19th 2009, 07:55 PM
justchillin
$a =\frac{dv}{dt}$