Question

a car accelerates at a constant 2 miles/minute. if i accelerate for a distance of 10 miles, how fast am I going?

Answer 1

i wonder if we integrate the accelration from 0 to 10 if we get the speed

Answer 2

i know if we integrate velocity we get displacement ... im just unsure if hte same thing applies here :)

Answer 3

\[\int 2 dt=2t\]
so velocity is: v(t) = 2t

but thats in time and not distance tho

Answer 4

distance would be the integration of velocity

\[\int 2t\ dt = t^2+t \]

Answer 5

when t^2 + t = 10 we should have the time :)

Answer 6

t^2+t=10

t^2+t+1/4 = 10 + 1/4

(t+1/2)^2 = 41/4

t = sqrt(41)/2 - 1/2

sounds resonable

Answer 7

v((sqrt(41)-1)/2) = sqrt(41) - 1

Answer 8

can't we use the formula s=ut+.5 a*t^2
so t=sqrt of 2s/a

Answer 9

dunno, that looks more like a physics abomination :)

Answer 10

yah, but it is applicable here as the acceleration is constant...

Answer 11

what the "u" turn out to be? I see that .5*2 = 1 is the same as mine

Answer 12

u means initial velocity here the body starts from rest so u=0..

Answer 13

so im prolly off on my ints ....

2

2t

t^2 + C ... yep