v(t) = 320(t+4)^(-1) is the velocity function, how do I find out the distance at 10 seconds?

Answer 1

distance is the derivative of velocity. so take the derivative of the function with respect to t, and then plug 10 in for t at the end.

v'(t)=-320(t+4)^-2
so your distance = -80/49 m (assuming that's your unit)

Answer 2

D(t) = 320 ∫ 1 / (t + 4) dt
D(t) = 320 ln(t + 4) + C
Assume at t = 0 , D = 0 and v is in m / sec
0 = 320 ln 4 + C
C = - 320 ln 4
D(t) = 320 [ ln(t + 4) - ln 4]
D(t) = 320 ln ((t + 4) / 4)
D(10) = 320 ln(14/4)
D(10) = 320 ln (7/2)
D(10) = 401 m

Answer 3

the last person is almost correct. You need to take the antiderivative and evaluate at 10. However to find the correct answer you need to solve this equation d(10) - d(0). That is the antiderivative evaluated at 10 minus the antiderivative evaluated at 0. This will give the correct distance.

Answer 4

Actually, velocity is the derivative of distance (not the other way around, as another person posted).

So, take the antiderivative of your velocity function (i.e. integrate), then solve for t=10.

btw, acceleration is the second derivative of distance (or first derivative of velocity).