# PhysicsMechanics- general motion in a straight line.

#### Shah 72

##### Member
A particle moves in a straight line. The velocity of the particle, v m/s, at time t s is given by v= -t^3+9t m/s for 0<t<5
a) Find the displacement of the particle from its original position, when t=5s
I got the ans for this by integration and limits 5 and 0 =- 43.8
b) work out the distance that the particle travels from t= 0 to t=5
I don't understand this. Velocity is positive from 0 to 3 and negative from 3 to 5 when I plot the velocity time graph.
I tried integration again with limits 3 to 0 and the next limit from 5 to 3. Iam not getting the ans which is 84.3m

#### skeeter

##### Well-known member
MHB Math Helper
in general, distance traveled is the integral of speed …

$\displaystyle D = \int_{t_0}^{t_f} |v(t)| \, dt$

Note the velocity in this problem is positive in the interval (0,3) and negative in the interval (3,5]

two ways to do this …

$\displaystyle D = \int_0^3 9t-t^3 \, dt + \int_3^5 t^3 - 9t \, dt$

$\displaystyle D = \int_0^3 9t-t^3 \, dt - \int_3^5 9t-t^3 \, dt$

#### Shah 72

##### Member
in general, distance traveled is the integral of speed …

$\displaystyle D = \int_{t_0}^{t_f} |v(t)| \, dt$

Note the velocity in this problem is positive in the interval (0,3) and negative in the interval (3,5]

two ways to do this …

$\displaystyle D = \int_0^3 9t-t^3 \, dt + \int_3^5 t^3 - 9t \, dt$

$\displaystyle D = \int_0^3 9t-t^3 \, dt - \int_3^5 9t-t^3 \, dt$
Thank you very much!