The motion of a particle is defined by the relation

x=2t^3−15t^2+24t+4 , where x and t are expressed in m and s. Determine where the velocity is zero and the position and total distance traveled when acceleration is zero.

This problem is really annoying! I have the answer to all the answers but the last one but the total distance traveled! Pls help!

## Answers (3)

If you throw a ball so that it reaches a height of 5m and you then catch it, you cannot determine the distance by finding x(0) = 0m and x(t) = 0m 0m-0m = 0m. The displacement was 0m, but the distance traveled was greater. If the final position were negative, that doesn't mean the distance traveled (a scalar) was negative.

You need to sum the positive distance traveled over all segments. Consider each segment to be separated by where v=0. In the ball example, one segment is from 0m to 5m (where v=0), then from 5m back to 0m. The total distance is 10m.

In this case we begin at t=0 in one direction until v=0 at t=1, then in another direction until a=0 at t=2.5

d(0-1) = x(1) - x(0) = 15 - 4 = 9

d(1-2.5) = x(1) - x(2.5) = 15 - 1.5 = 13.5

d(0-2.5) = 9 +13.5 = 22.5

EDIT:

" I thought it would be the integral of velocity from 0 to 2.5"

It's the integral of the absolute value of velocity from 0 to 2.5

The velocity equation is the 1st derivative of the distance equation.

v = 6t^2 – 30t + 24

t = [30 Â± â(900 – 576) Ã· 6

t = [30 Â± 18] Ã· 6

t = 8 seconds and 2 seconds

Now we can determine 2 positions.

x= 2 * 8^3 â 15 * 8^2 + 24 * 8 + 4 = 1024 – 960 + 192 + 4 = 260 m

x= 2 * 2^3 â 15 * 2^2 + 24 * 2 + 4 = 16 – 60 + 48 + 4 = 8 m

The acceleration equation is the 1st derivative of the velocity equation.

v = 6t^2 – 30t + 24

a = 12 t – 30 = 0

t = 2.5 seconds

x= 2 * 2.5^3 â 15 * 2.5^2 + 24 * 2.5 + 4 = 31.25 – 93.75 + 60 + 4 = 1.5 m

v = x' = 6t^2 - 30t + 24

a = v' = 12*t - 30

v = 0 = 6t^2 - 30t + 24 = 6*(t^2 - 5t + 4) = 6*(t - 4)(t - 1)

0 = 6*(t-4)*(t-1) ==> t = 4 and t = 1

Plug those numbers back into the x(t) equation to find the position.

a = 0 = 12*t - 30 ==> t = 2.5

Total distance traveled = h(2.5) - h(0)